problem solving free falling bodies

3 Motion Along a Straight Line

3.5 free fall, learning objectives.

By the end of this section, you will be able to:

• Use the kinematic equations with the variables y and g to analyze free-fall motion.
• Describe how the values of the position, velocity, and acceleration change during a free fall.
• Solve for the position, velocity, and acceleration as functions of time when an object is in a free fall.

An interesting application of (Figure) through (Figure) is called free fall , which describes the motion of an object falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size. Let’s assume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. But “falling,” in the context of free fall, does not necessarily imply the body is moving from a greater height to a lesser height. If a ball is thrown upward, the equations of free fall apply equally to its ascent as well as its descent.

The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration , independent of their mass . This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones. Until Galileo Galilei (1564–1642) proved otherwise, people believed that a heavier object has a greater acceleration in a free fall. We now know this is not the case. In the absence of air resistance, heavy objects arrive at the ground at the same time as lighter objects when dropped from the same height (Figure) .

Figure 3.26 A hammer and a feather fall with the same constant acceleration if air resistance is negligible. This is a general characteristic of gravity not unique to Earth, as astronaut David R. Scott demonstrated in 1971 on the Moon, where the acceleration from gravity is only 1.67 m/s2 and there is no atmosphere.

In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball reaches the ground after a baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, and friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them.

For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free fall . The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called acceleration due to gravity . Acceleration due to gravity is constant, which means we can apply the kinematic equations to any falling object where air resistance and friction are negligible. This opens to us a broad class of interesting situations.

Acceleration due to gravity is so important that its magnitude is given its own symbol, g . It is constant at any given location on Earth and has the average value

Although g varies from 9.78 m/s 2 to 9.83 m/s 2 , depending on latitude, altitude, underlying geological formations, and local topography, let’s use an average value of 9.8 m/s 2 rounded to two significant figures in this text unless specified otherwise. Neglecting these effects on the value of g as a result of position on Earth’s surface, as well as effects resulting from Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth). In fact, its direction defines what we call vertical. Note that whether acceleration a in the kinematic equations has the value + g or − g depends on how we define our coordinate system. If we define the upward direction as positive, then [latex] a=\text{−}g=-9.8\,{\text{m/s}}^{2}, [/latex] and if we define the downward direction as positive, then [latex] a=g=9.8\,{\text{m/s}}^{2} [/latex].

One-Dimensional Motion Involving Gravity

The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So, we start by considering straight up-and-down motion with no air resistance or friction. These assumptions mean the velocity (if there is any) is vertical. If an object is dropped, we know the initial velocity is zero when in free fall. When the object has left contact with whatever held or threw it, the object is in free fall. When the object is thrown, it has the same initial speed in free fall as it did before it was released. When the object comes in contact with the ground or any other object, it is no longer in free fall and its acceleration of g is no longer valid. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g . We represent vertical displacement with the symbol y .

Kinematic Equations for Objects in Free Fall

We assume here that acceleration equals − g (with the positive direction upward).

Problem-Solving Strategy: Free Fall

• Decide on the sign of the acceleration of gravity. In (Figure) through (Figure) , acceleration g is negative, which says the positive direction is upward and the negative direction is downward. In some problems, it may be useful to have acceleration g as positive, indicating the positive direction is downward.
• Draw a sketch of the problem. This helps visualize the physics involved.
• Record the knowns and unknowns from the problem description. This helps devise a strategy for selecting the appropriate equations to solve the problem.
• Decide which of (Figure) through (Figure) are to be used to solve for the unknowns.

Free Fall of a Ball (Figure) shows the positions of a ball, at 1-s intervals, with an initial velocity of 4.9 m/s downward, that is thrown from the top of a 98-m-high building. (a) How much time elapses before the ball reaches the ground? (b) What is the velocity when it arrives at the ground?

Figure 3.27 The positions and velocities at 1-s intervals of a ball thrown downward from a tall building at 4.9 m/s.

Choose the origin at the top of the building with the positive direction upward and the negative direction downward. To find the time when the position is −98 m, we use (Figure) , with [latex] {y}_{0}=0,{v}_{0}=-4.9\,\text{m/s,}\,\text{and}\,g=9.8\,{\text{m/s}}^{2} [/latex].

• Show Answer Substitute the given values into the equation: [latex] \begin{array}{cc} y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}\hfill \\ -98.0\,\text{m}=0-(4.9\,\text{m/s})t-\frac{1}{2}(9.8\,{\text{m/s}}^{2}){t}^{2}.\hfill \end{array} [/latex] This simplifies to [latex] {t}^{2}+t-20=0. [/latex] This is a quadratic equation with roots [latex] t=-5.0\mathrm{s}\,\text{and}\,t=4.0\mathrm{s} [/latex]. The positive root is the one we are interested in, since time [latex] t=0 [/latex] is the time when the ball is released at the top of the building. (The time [latex] t=-5.0\mathrm{s} [/latex] represents the fact that a ball thrown upward from the ground would have been in the air for 5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)
• Show Answer Using (Figure), we have [latex] v={v}_{0}-gt=-4.9\,\text{m/s}-(9.8{\text{m/s}}^{2})(4.0\,\text{s})=-44.1\,\text{m/s}\text{.} [/latex]

Significance

For situations when two roots are obtained from a quadratic equation in the time variable, we must look at the physical significance of both roots to determine which is correct. Since [latex] t=0 [/latex] corresponds to the time when the ball was released, the negative root would correspond to a time before the ball was released, which is not physically meaningful. When the ball hits the ground, its velocity is not immediately zero, but as soon as the ball interacts with the ground, its acceleration is not g and it accelerates with a different value over a short time to zero velocity. This problem shows how important it is to establish the correct coordinate system and to keep the signs of g in the kinematic equations consistent.

Vertical Motion of a Baseball

A batter hits a baseball straight upward at home plate and the ball is caught 5.0 s after it is struck (Figure) . (a) What is the initial velocity of the ball? (b) What is the maximum height the ball reaches? (c) How long does it take to reach the maximum height? (d) What is the acceleration at the top of its path? (e) What is the velocity of the ball when it is caught? Assume the ball is hit and caught at the same location.

Figure 3.28 A baseball hit straight up is caught by the catcher 5.0 s later.

Choose a coordinate system with a positive y -axis that is straight up and with an origin that is at the spot where the ball is hit and caught.

• Show Answer The acceleration is 9.8 m/s2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s2 downward.
• Show Answer The velocity at [latex] t=5.0\mathrm{s} [/latex] can be determined with (Figure): [latex] \begin{array}{cc}\hfill v& ={v}_{0}-gt\hfill \\ & =24.5\,\text{m/s}-9.8{\text{m/s}}^{2}(5.0\,\text{s})\hfill \\ & =-24.5\,\text{m/s}.\hfill \end{array} [/latex]

The ball returns with the speed it had when it left. This is a general property of free fall for any initial velocity. We used a single equation to go from throw to catch, and did not have to break the motion into two segments, upward and downward. We are used to thinking of the effect of gravity is to create free fall downward toward Earth. It is important to understand, as illustrated in this example, that objects moving upward away from Earth are also in a state of free fall.

Check Your Understanding

A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Which quantity increases faster, the speed of the ice chunk or its distance traveled?

It takes 2.47 s to hit the water. The quantity distance traveled increases faster.

Rocket Booster

A small rocket with a booster blasts off and heads straight upward. When at a height of [latex] 5.0\,\text{km} [/latex] and velocity of 200.0 m/s, it releases its booster. (a) What is the maximum height the booster attains? (b) What is the velocity of the booster at a height of 6.0 km? Neglect air resistance.

Figure 3.29 A rocket releases its booster at a given height and velocity. How high and how fast does the booster go?

We need to select the coordinate system for the acceleration of gravity, which we take as negative downward. We are given the initial velocity of the booster and its height. We consider the point of release as the origin. We know the velocity is zero at the maximum position within the acceleration interval; thus, the velocity of the booster is zero at its maximum height, so we can use this information as well. From these observations, we use (Figure) , which gives us the maximum height of the booster. We also use (Figure) to give the velocity at 6.0 km. The initial velocity of the booster is 200.0 m/s.

• From (Figure) , Show Answer [latex] {v}^{2}={v}_{0}^{2}-2g(y-{y}_{0}) [/latex]. With [latex] v=0\,\text{and}\,{y}_{0}=0 [/latex], we can solve for y: [latex] y=\frac{{v}_{0}^{2}}{-2g}=\frac{(2.0\,×\,{10}^{2}\text{m}\text{/}{\text{s})}^{2}}{-2(9.8\,\text{m}\text{/}{\text{s}}^{2})}=2040.8\,\text{m}\text{.} [/latex] This solution gives the maximum height of the booster in our coordinate system, which has its origin at the point of release, so the maximum height of the booster is roughly 7.0 km.

We have, from (Figure) ,

We have both a positive and negative solution in (b). Since our coordinate system has the positive direction upward, the +142.8 m/s corresponds to a positive upward velocity at 6000 m during the upward leg of the trajectory of the booster. The value v = −142.8 m/s corresponds to the velocity at 6000 m on the downward leg. This example is also important in that an object is given an initial velocity at the origin of our coordinate system, but the origin is at an altitude above the surface of Earth, which must be taken into account when forming the solution.

Visit this site to learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (for example, y = bx ) to see how they add to generate the polynomial curve.

• An object in free fall experiences constant acceleration if air resistance is negligible.
• On Earth, all free-falling objects have an acceleration g due to gravity, which averages [latex] g=9.81\,{\text{m/s}}^{2} [/latex].
• For objects in free fall, the upward direction is normally taken as positive for displacement, velocity, and acceleration.

Conceptual Questions

What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down? Assume there is no air resistance.

An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Does its velocity change direction? (c) Does the acceleration have the same sign on the way up as on the way down?

a. at the top of its trajectory; b. yes, at the top of its trajectory; c. yes

Suppose you throw a rock nearly straight up at a coconut in a palm tree and the rock just misses the coconut on the way up but hits the coconut on the way down. Neglecting air resistance and the slight horizontal variation in motion to account for the hit and miss of the coconut, how does the speed of the rock when it hits the coconut on the way down compare with what it would have been if it had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way up or down? Explain.

The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the same, how many times higher could a safe fall on the Moon than on Earth (gravitational acceleration on the Moon is about one-sixth that of the Earth)?

Earth [latex] v={v}_{0}-gt=\text{−}gt [/latex]; Moon [latex] {v}^{\prime }=\frac{g}{6}{t}^{\prime }v={v}^{\prime }\enspace-gt=-\frac{g}{6}{t}^{\prime }\enspace{t}^{\prime }=6t [/latex]; Earth [latex] y=-\frac{1}{2}g{t}^{2} [/latex] Moon [latex] {y}^{\prime }=-\frac{1}{2}\,\frac{g}{6}{(6t)}^{2}=-\frac{1}{2}g6{t}^{2}=-6(\frac{1}{2}g{t}^{2})=-6y [/latex]

How many times higher could an astronaut jump on the Moon than on Earth if her takeoff speed is the same in both locations (gravitational acceleration on the Moon is about on-sixth of that on Earth)?

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be [latex] {y}_{0}=0 [/latex].

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

a. [latex] \begin{array}{cc} y=-8.23\,\text{m}\hfill \\ {v}_{1}=\text{−}18.9\,\text{m/s}\hfill \end{array} [/latex];

b. [latex] \begin{array}{cc} y=-18.9\,\text{m}\hfill \\ {v}_{2}=23.8\,\text{m/s}\hfill \end{array} [/latex];

c. [latex] \begin{array}{cc} y=-32.0\,\text{m}\hfill \\ {v}_{3}=\text{−}28.7\,\text{m/s}\hfill \end{array} [/latex];

d. [latex] \begin{array}{cc} y=-47.6\,\text{m}\hfill \\ {v}_{4}=\text{−}33.6\,\text{m/s}\hfill \end{array} [/latex];

e. [latex] \begin{array}{cc} y=-65.6\,\text{m}\hfill \\ {v}_{5}=\text{−}38.5\,\text{m/s}\hfill \end{array} [/latex]

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.

a. Knowns: [latex] a=\text{−}9.8\,{\text{m/s}}^{2}\enspace{v}_{0}=-1.4\,\text{m/s}t=1.8\,\text{s}\enspace{y}_{0}=0\,\text{m} [/latex];

b. [latex] y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}y={v}_{0}t-\frac{1}{2}gt=-1.4\,\text{m}\text{/}\text{s}(1.8\,\text{sec})-\frac{1}{2}(9.8){(1.8\,\text{s})}^{2}=-18.4\,\text{m} [/latex] and the origin is at the rescuers, who are 18.4 m above the water.

Unreasonable results A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known, and identify its value. Then, identify the unknown and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long a time is the dolphin in the air? Neglect any effects resulting from his size or orientation.

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

a. [latex] {v}^{2}={v}_{0}^{2}-2g(y-{y}_{0})\enspace{y}_{0}=0v=0y=\frac{{v}_{0}^{2}}{2g}=\frac{{(4.0\,\text{m}\text{/}\text{s})}^{2}}{2(9.80)}=0.82\,\text{m} [/latex]; b. to the apex [latex] v=0.41\,\text{s} [/latex] times 2 to the board = 0.82 s from the board to the water [latex] y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}y=-1.80\,\text{m}\enspace{y}_{0}=0\enspace{v}_{0}=4.0\,\text{m}\text{/}\text{s} [/latex][latex] -1.8=4.0t-4.9{t}^{2}\enspace4.9{t}^{2}-4.0t-1.80=0 [/latex], solution to quadratic equation gives 1.13 s; c. [latex] \begin{array}{c}{v}^{2}={v}_{0}^{2}-2g(y-{y}_{0})\enspace{y}_{0}=0\enspace\enspace{v}_{0}=4.0\,\text{m}\text{/}\text{s}y=-1.80\,\text{m}\hfill \\ v=7.16\,\text{m}\text{/}\text{s}\hfill \end{array} [/latex]

(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long a time would it take to reach the ground if it is thrown straight down with the same speed?

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long a time does he have to get out of the way if the shot was released at a height of 2.20 m and he is 1.80 m tall?

Time to the apex: [latex] t=1.12\,\text{s} [/latex] times 2 equals 2.24 s to a height of 2.20 m. To 1.80 m in height is an additional 0.40 m. [latex] \begin{array}{cc} y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}y=-0.40\,\text{m}\enspace{y}_{0}=0\enspace{v}_{0}=-11.0\,\text{m}\text{/}\text{s}\hfill \\ y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}y=-0.40\,\text{m}\enspace{y}_{0}=0\enspace{v}_{0}=-11.0\,\text{m}\text{/}\text{s}\hfill \\ -0.40=-11.0t-4.9{t}^{2}\enspace\text{or}\enspace4.9{t}^{2}+11.0t-0.40=0\hfill \end{array} [/latex].

Take the positive root, so the time to go the additional 0.4 m is 0.04 s. Total time is [latex] 2.24\,\text{s}\,+0.04\,\text{s}\,=2.28\,\text{s} [/latex].

You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.0 m. How much additional time elapses before the ball passes the tree branch on the way back down?

A kangaroo can jump over an object 2.50 m high. (a) Considering just its vertical motion, calculate its vertical speed when it leaves the ground. (b) How long a time is it in the air?

a. [latex] \begin{array}{cc} {v}^{2}={v}_{0}^{2}-2g(y-{y}_{0})\enspace{y}_{0}=0v=0y=2.50\,\text{m}\hfill \\ {v}_{0}^{2}=2gy⇒{v}_{0}=\sqrt{2(9.80)(2.50)}=7.0\,\text{m}\text{/}\text{s}\hfill \end{array} [/latex]; b. [latex] t=0.72\,\text{s} [/latex] times 2 gives 1.44 s in the air

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105.0 m. He can’t see the rock right away, but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long a time will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335.0 m/s on this day.

a. [latex] v=70.0\,\text{m}\text{/}\text{s} [/latex]; b. time heard after rock begins to fall: 0.75 s, time to reach the ground: 6.09 s

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Free Fall Motion: Explanation, Review, and Examples

• The Albert Team
• Last Updated On: February 16, 2023

Free fall and projectile motion describe objects that are moving through the air and acted on only by gravity. In this post, we will describe this type of motion using both graphs and kinematic equations. Since projectile motion involves two dimensions, these problems can be complex. We will explain many examples so you can see how to solve different types of projectile motion. 

What We Review

An object that is moving under only the influence of gravity is in free fall. In order for an object to be in free fall, wind and air resistance must be ignored. On Earth, all objects in free fall accelerate downward at the rate of gravity or 9.81\text{ m/s}^2 .

Applying Free Fall to Kinematic Equations

When analyzing free fall motion, we can apply the same kinematic equations as we did for motion on the ground. We can then use these equations to determine properties such as distance, time, and velocity. 

How to Find Distance Fallen for an Object in Free Fall

If an object is in free fall, we can use kinematic equations to find the distance it falls during a certain time. You will typically use the following kinematic equation to calculate the distance fallen:

In order to use this equation, you need to know the initial velocity of the object and the time of flight. Remember that the acceleration of a free falling object is always equal to the acceleration due to gravity, 9.81\text{ m/s}^2 . 

Many free fall physics problems will include scenarios where objects are dropped from rest. In this case, the initial velocity is zero and the first term of the kinematic equation above will cancel out. 

If the time is not known, another method for calculating the distance fallen is to use the following kinematic equation:

In this case, you must know the final velocity v_f of the object. Then, you can solve the equation for the distance d .

How to Find Time for an Object in Free Fall

The amount of time an object is in free fall will depend on its velocity and the distance it falls. Similar to distance, there are two equations you can use to find the time, depending on what you know. 

If you know the initial and final velocity of the object, then the simplest way to calculate time is using the kinematic equation:

This equation can be solved for time. Then, you’ll only need to substitute the values for the velocities and the acceleration due to gravity.

Another method to find time if you do not know the object’s final velocity is to use the equation:

Note that in this equation there are two terms that include the time t . Unless the initial velocity is zero, this can make it more challenging to solve this equation for time. If using this equation, you may need to use the quadratic formula to solve for time.

How to Find Final Velocity for an Object in Free Fall

The final velocity of an object in free fall depends on the amount of time it falls. Due to the acceleration of gravity, the velocity will increase every second by 9.81\text{ m/s} . The final velocity can be calculated using the equation:

If you do not know the amount of time the object is falling, another method for calculating the final velocity is using the kinematic equation: 

This equation requires that you instead know the distance that the object falls. If you are using this equation to find the final velocity, remember that the final velocity is squared in this equation. That means you will need to take a square root as your final step to solve for the final velocity. 

Examples of Free Fall

In this next section, we’ll apply the methods you just learned to solve some problems about free fall motion.

Example 1: How to Find the Distance for an Object Dropped from Rest

For example, an object is dropped from rest from the top of a tall building. It hits the ground 5\text{ s} after it is dropped. What is the height of the building? 

In this scenario, we know that the object’s initial velocity is zero because it was dropped from rest. We also know that the acceleration is 9.81\text{ m/s}^2 . This problem is asking us to find the distance the object falls. This will be equal to the height of the building.

Based on this information, we can use the following kinematic equation to find the distance:

Substituting the given values produces:

Therefore, the height of the building is about 123\text{ m} .

Example 2: How to Find the Final Velocity for an Object with Initial Velocity

In another example, an object in free fall has an initial, downward velocity of 2\text{ m/s} and falls a distance of 45\text{ m} . What is the object’s final velocity? 

In this scenario, we are given the object’s initial velocity, v_i and the distance d . We also know that the acceleration is 9.81\text{ m/s}^2 . Based on this information, we can use the following kinematic equation to find the final velocity:

Since the initial velocity is in the same direction as the acceleration (downward) we can use the same sign for both values.

Our last step is to eliminate the square by taking the square root:

Therefore, the final velocity of the object is about 30\text{ m/s} .

Motion Graphs for Objects in Free Fall

In addition to using physics equations, we can also represent free fall motion with motion graphs. Position-time graphs, velocity-time graphs, and acceleration-time graphs can tell us a lot about the object’s motion over time. Want a more in-depth review of motion graphs? Check out this blog post !

Position-Time Graph for an Object in Free Fall

In terms of position, many objects in free fall start at a high position, or height off the ground, and move downward. Objects in free fall accelerate due to gravity. Therefore, the position-time graph for free fall motion must be curved. This means that objects in free fall start with a slow velocity and gradually speed up which is represented by the steep downward curve of the graph. 

Velocity-Time Graph for an Object in Free Fall

As an object falls, its velocity increases due to the acceleration of gravity. This means that the velocity starts slow and steadily increases in the downward direction. The graph below shows the velocity-time for an object in free fall:

Note that the slope of this graph is constant and represents the acceleration due to gravity, or -9.81\text{ m/s}^2 .

Acceleration-Time Graph for an Object in Free Fall

Free fall acceleration is constant. Throughout the entire time that an object is falling, it is accelerating at a rate equal to the acceleration due to gravity, -9.81\text{ m/s}^2 . As shown in the graph below, the acceleration-time graph is a constant negative line. 

Projectile Motion

A projectile is an object that is launched or thrown into the air and then only influenced by gravity. Projectile motion has many similarities to free fall motion, however, projectiles may also travel a horizontal distance in addition to falling vertically down. 

Examples of Projectile Motion

The exact trajectory, or path, a projectile will take depends on how it is launched. However, all projectiles follow a curved trajectory such as in the image shown below:

If you play or watch sports, you likely have already observed projectile motion. Projectile motion describes the arc of a basketball in a free throw, a fly ball in baseball, or a volleyball bumped over the net. 

Horizontal Component of Velocity

To analyze projectile motion, we must separate the motion into horizontal and vertical components. The horizontal component of a projectile’s velocity is independent of the vertical component of velocity. Since gravity acts vertically, there are no horizontal forces acting on projectiles. This means that the horizontal component of a projectile’s velocity remains constant throughout the entire flight. 

Example: Finding the Horizontal Component

For example, a projectile is launched from the ground with an initial speed of 8\text{ m/s} at a 60^{\circ} angle. What is the horizontal component of the projectile’s velocity?

We will need to use trig identities to determine the components of the velocity. We can visualize the components as a triangle where the hypotenuse is the initial velocity and the sides represent the horizontal, v_{ix} , and vertical, v_{iy} , components of the velocity.

Cosine is defined as the adjacent side of the triangle divided by the hypotenuse. Since the horizontal component is adjacent to the angle, we can use cosine to find the horizontal component of velocity:

Therefore, the horizontal component of the initial velocity is 4\text{ m/s} .

Need to review your trig identities? Try out this resource from Khan Academy .

Vertical Component of Velocity

The vertical component of a projectile’s velocity will be influenced by gravity, which acts vertically on the object causing it to accelerate downward. Therefore, the vertical component of velocity will change throughout the projectile’s flight. We can calculate the vertical component of velocity at a particular time in a method similar to calculating the horizontal component. 

Example: Finding the Vertical Component

In the same example, a projectile is launched from the ground with an initial speed of 8\text{ m/s} at a 60^{\circ} angle. What is the vertical component of the projectile’s velocity?

As we visualize the velocity components, we are solving this time for the opposite side of the triangle. Sine is defined as the opposite side of the triangle divided by the hypotenuse. Therefore, the initial vertical velocity is:

Solving Projectile Motion Questions

Let’s apply what we’ve learned to some examples of projectile motion!

Example 1: Finding the Range of a Projectile

In this example, a projectile is fired horizontally with a speed of 5\text{ m/s} from a cliff with a height of 60\text{ m} . How far from the base of the cliff will the projectile land? 

In this scenario, we are given the initial horizontal velocity v_{ix}=5\text{ m/s} and the vertical change in position d_y=-60\text{ m} . Since the projectile is launched horizontally, the initial vertical velocity, v_{iy} , is zero. We also always know in projectile motion that the vertical acceleration is a_y=-9.81\text{ m/s}^2 and the horizontal acceleration, a_x , is zero.

This problem is asking us to find the horizontal displacement, or d_x . This is also referred to as the range . We can use the following kinematic equation to find the projectile’s final horizontal position:

Since the horizontal acceleration of a projectile is zero, this equation can be simplified to:

Before we can solve this equation, we must first determine the time of the projectile’s flight. We can actually use this same equation in the vertical direction to solve for time:

Since the initial vertical velocity is zero, this equation can be simplified to:

Solving for t :

Substituting the given values:

Now we can use this time to calculate the horizontal displacement of the projectile:

Therefore, the projectile will land about 17.5\text{ m} from the base of the cliff. 

Example 2: Finding the Maximum Height of a Projectile

As another example, a projectile is launched from the ground with an initial velocity of 25\text{ m/s} at an angle of 50^{\circ} . What is the projectile’s maximum height?

As a projectile travels upward, its vertical velocity becomes slower and slower due to the negative acceleration of gravity. At the maximum height of the trajectory, the projectile’s vertical velocity will momentarily be zero as the projectile stops and turns to move downward. Therefore, in this scenario, our final vertical velocity, v_{fy} , is zero.

We can use the following kinematic equation to solve for the maximum height, d_y :

Solving for d_y :

Before we can use this equation to calculate the height, we will need to use the sine trig identity to find the vertical component of the initial velocity:

Since the initial velocity is in the opposite direction as the acceleration, it’s really important to remember the sign here. If we define moving up as positive, then the initial velocity is positive and the acceleration is negative. Substituting this initial vertical velocity and the given values into the equation above gives:

Therefore, the projectile will reach a maximum height of about 18.7\text{ m} .

For more examples and an explanation of solving these types of projectile motion problems, check out this youtube video from Professor Dave . 

Understanding free fall and projectile motion allows you to solve some of the most complex problems you will encounter in introductory physics. All projectiles are acted on only by gravity, and the vertical and horizontal components of motion are independent of each other. This allows us to apply our kinematic equations to solve for a projectile’s time of flight, velocity, and displacement in each direction.  

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Free Fall Practice Problems for High Schools: Complete Guide

In this long article, we are going to practice some problems about a freely falling object in the absence of air resistance. All these questions are suitable for high school or college students, or even the AP Physics 1 exam. 

Freely Falling Motion Problems

Problem (1): A tennis ball is thrown vertically upward with an initial speed of 17 m/s and caught at the same level above the ground.  (a) How high does the ball rise? (b) How long was the ball in the air? (c) How long does it take to reach its highest point?

Solution : Take up as the positive direction and the throwing point as the origin, so $y_0=0$. 

(a) The ball goes up so high that its vertical velocity becomes zero. For this part of ascending motion, we can use the free fall kinematic equation $v^2-v_0^2=-2g(y-y_0)$. Substituting the known values into it and solving for $y$, we get \begin{gather*} v^2-v_0^2=-2g(y-y_0) \\\\ 0-17^2 = -2(10)(y_{max}-0) \\\\ \Rightarrow \boxed{y_{max}=14.45\,\rm m} \end{gather*}  (b) In all free fall practice problems, the best way to find the total flight time that the object was in the air is to use the kinematic equation $y-y_0=-\frac 12 gt^2+v_0t$. Then, substitute the coordinates of where the object landed on the ground into it. 

As a rule of thumb, if the object returns to the same point of launch, its displacement vector is always zero, so $y-y_0=0$. Therefore, we have \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ 0=-\frac 12 (10)t^2+17t \\\\ \Rightarrow \boxed{5t^2-17t=0} \end{gather*} Solving this equation by factoring out the time and setting the remaining expression to zero, we get  \begin{gather*} 5t^2-17t=0 \\\\ t(5t-17)=0 \\\\ \Rightarrow t=0 \quad ,\quad t=3.4\,\rm s \end{gather*} The first result corresponds to the initial time, and the other time, $\boxed{t_{tot}=3.4\,\rm s}$, is the amount of time the ball is in the air until it reaches the ground. 

(c) At the highest point the vertical velocity is always zero, $v=0$. Using the equation $v=v_0-gt$, and solving for $t$, we have \begin{gather*} v=v_0-gt \\ 0=17-(10)t \\ \Rightarrow t_{top}=1.7\,\rm s \end{gather*} As you can see, the duration of the ball's going up, in the absence of air resistance, is always half the total flight time. Therefore, \[t_{top}=\frac 12 t_{tot}\]

Problem (2): A ball is dropped directly downward from a height of 45 meters with an initial speed of 6 m/s. How many seconds later does it strike the ground?

Solution : Taking up as the positive direction and the dropping point as the origin, we have $y_0=0$. Since the ball is moving downward, we choose a negative sign for its initial velocity, so $v_0=-6\,\rm m/s$.

The ball strikes the ground $45\,\rm m$ below the chosen origin, so its correct coordinate is $y=-45\,\rm m$. 

The only kinematic equation that relates all these variables to the time is $y-y_0=-\frac 12 gt^2+v_0t$. Substituting the numerical values into this equation, yields \begin{gather*} y-y_0 =-\frac 12 gt^2+v_0t \\\\ -45-0 =-\frac 12 (10)t^2+(-6)t \\\\ \Rightarrow \boxed{5t^2+6t-45=0} \end{gather*} In the last step, after rearranging, we arrived at a quadratic equation, like $at^2+bt+c=0$, that its solution is found using the below formula \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \] where $a,b,c$ are constants. In this case, we have \[a=5, b=5, c=-45\] Substituting these values into the above formula, we get \begin{gather*} t=\frac{-5\pm\sqrt{5^2-4(5)(-45)}}{2(5)} \\\\ t=2.46\,\rm s \quad , \quad t=-3.66 \end{gather*} Since we choose the positive time for free fall problems, the ball would reach the ground approximately $2.5\,\rm s$ after being dropped.

Problem (3): A coin is tossed vertically upward and remains $\rm 0.6\, s$ in the air before it comes back down.  (a) How fast is the coin tossed? (b) How high does the coin rise? 

Solution : Let the tossing point be the origin, so $y_0=0$. The time duration the coin is in the air until it reaches the highest point is $t=0.6\,\rm s$. Recall that in all free fall problems, the object at the highest point has a velocity $v=0$. 

(a) Use the kinematic equation $v=v_0-gt$, substitute the above-known values, and solve for the unknown initial speed $v_0$ \begin{align*} v&=v_0-gt \\ 0&=v_0-(10)(0.6) \\ \Rightarrow v_0&=6\,\rm m/s \end{align*} Therefore, the coin was tossed vertically upward with an initial speed of $6\,\rm m/s$. 

(b) In this part, we need a kinematic equation that relates the distance traveled to the time taken, $y-y_0=-\frac 12 gt^2+v_0t$, or to the initial and final velocities, $v^2-v_0^2=-2g(y-y_0)$. We want to use the second equation as follows. \begin{align*} v^2-v_0^2&=-2g(y-y_0) \\\\ (0)^2-(60)^2 &= -2(10)(y-0) \\\\ y&=\frac{-6^2}{-20} \\\\ &=\boxed{1.8\,\rm m} \end{align*} You can also use the first equation and arrive at the same result. Check it out yourself. 

Problem (4): A small stone is shot straight up with an initial speed of $\rm 15\,m/s$.  (a) How long does the stone take to reach its maximum height? (b) How high does the stone go? (c) With what speed would the stone hit the ground?

Solution : The known data is $y_0=0$ and $v_0=15\,\rm m/s$. 

(a) When an object is thrown vertically upward in the air, it rises until it reaches a point where its vertical velocity becomes zero; otherwise, it continues on its way. So, in this part, at the maximum height, we have $v=0$. Substituting all these known numerical values into the equation $v=v_0-gt$, and solving for the unknown time $t$, we get \begin{align*} 0 &= 15-(10)t \\ \Rightarrow t&=\boxed{1.5\,\rm s} \end{align*} Thus, it takes $1.5\,\rm s$ for  the stone to reach the highest point. 

(b) For this part, we have the time taken to reach the maximum height $t=1.5\,\rm s$, the initial and final velocities. Thus, we can use either the equation $y-y_0=-\frac 12 gt^2+v_0t$ or $v^2-v_0^2=-2g(y-y_0)$. We chose the second equation. \begin{align*} 0-(15)^2 &= -2(10)(y_{max}-0) \\\\ y_{max} &=\frac{-15^2}{-20} \\\\ \Rightarrow y_{max}&= \boxed{11.25\,\rm m} \end{align*} Thus, the stone reaches the maximum height of $11.25\,\rm m$. 

(c) When the stone hits the ground at the same level as the throwing point, then according to the definition of displacement, it displaces nothing. This means that the displacement between initial and final points $\Delta y=y-y_0$ is zero. 

Using this fact, we can use the equation $v^2-v_0^2=-2g(y-y_0)$ to get the velocity at the moment of hitting the ground. \begin{align*} v^2-(15)^2 &=-2(10)(0) \\ \Rightarrow v^2 = 15^2 \end{align*} Taking the square root of both sides, yields two roots of $v=\pm 15\,\rm m/s$. Recall that velocity is a vector in physics that has both magnitude and direction. 

The stone is hitting the ground, so the correct sign is a negative for its velocity, i.e., $v=-15\,\rm m/s$. 

Problem (5): From the top of a high cliff, a stone is released. It is seen after $2.5\,\rm s$, the stone strikes the ground. How high is the cliff? 

Solution : The stone is released, so its initial speed is zero, $v_0=0$. The total flight time is also $t=2.5\,\rm s$. Taking the releasing point as the origin, we will have $y_0=0$. In the kinematic equations, there is vertical displacement in only two of them, i.e., $v^2-v_0^2=-2g(y-y_0)$ and $y-y_0=-\frac 12 gt^2+v_0t$. As you can see, to use the first equation, we must have the final velocity where the stone hits the ground, and for the second equation, the total flight time is needed. Hence, it is simpler to apply the second equation and solve for the unknown vertical distance $y$.  \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ y-0=-\frac 12 (10)(2.5)^2+(0)(2.5) \\\\ \Rightarrow \quad \boxed{y=-31.25\,\rm m} \end{gather*} The negative indicates that the stone strikes the ground $31.25\,\rm m$ below our chosen origin. 

Problem (6): A stone is released at rest from a height and falls freely for $4\,\rm s$.  (a) What is the stone's velocity $1.5\,\rm s$ after releasing? (b) How high is the height?

Solution : The stone is released at rest, so $v_0=0$. Take upward as the positive direction. The total flight time is $t_{tot}=4\,\rm s$. 

(a) With this known information, use the equation $v=v_0-gt$ to find the stone's velocity at each instant of time. \begin{gather*} v=v_0-gt \\ v=0-(10)(1.5) \\ \Rightarrow \boxed{v=-15\,\rm m/s} \end{gather*} The negative indicates the direction of the stone at that moment, which is facing down. 

(b) The time between releasing the stone and hitting the ground is given. With this known information, use the equation $y-y_0=-\frac 12 gt^2 +v_0t$ and solve for the total distance fallen by the stone. \begin{gather*} y-0=-\frac 12 (10)(4)^2+(0)(4) \\\\ \Rightarrow \quad \boxed {y=-80\,\rm m} \end{gather*} The minus sign reminds us that the stone strikes the ground $80\,\rm m$ below our chosen origin. 

Problem (7): A person throws a light stone straight up and catches it $2.6\,\rm s$ later. With what speed did he throw the stone, and to what height does the stone go up? 

Solution : As with any other free-fall problem, let upward be the positive direction and the throwing point be the origin of the coordinate system so that $y_0=0$. 

The stone is caught at the same level of throwing, so its vertical displacement is zero, $\Delta y=y-y_0=0$. The total flight time is also known. Hence, applying the vertical displacement kinematic equation, $y-y_0=-\frac 12 gt^2+v_0t$, and solving for the initial speed $v_0$ gives us \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ 0=-\frac 12 (10)(2.6)^2+v_0 (2.6) \end{gather*} Factoring out $2.6$ from the last expression, would get \begin{gather*} (2.6)(-5(2.6)+v_0)=0 \\\\ \Rightarrow \quad \boxed{v_0=13\,\rm m/s} \end{gather*} As a side note, if an object is thrown vertically upward and caught at the same level, by knowing the time interval between these two moments, we can use the following formula to find its initial speed \[v_0=\frac 12 g t\] The stone goes up until it reaches a point where its vertical velocity is zero, i.e., $v=0$. Now that we know the stone's initial speed, applying the equation $v^2-v_0^2=-2g(y-y_0)$ gives us \begin{gather*} 0-(13)^2=-2(10)(y-0) \\\\ \Rightarrow \quad \boxed{y=8.45\,\rm m} \end{gather*} 

Problem (8): A baseball is thrown straight up with a speed of $25\,\rm m/s$.  (a) With what speed is it moving when it is at a height of $10\,\rm m$? (b) How much time does it take to reach that point?

Solution : As usual, take up the positive $y$-direction and set $y_0=0$. The initial speed is also $v_0=25\,\rm m/s$. 

(a) The only time-independent kinematic equation that relates these known values to each other is $v^2-v_0^2=-2g(y-y_0)$. Substituting the known numerical values into this, we get \begin{gather*} v^2-(25)^2=-2(9.8)(10-0) \\\\ v^2=429 \\\\ \Rightarrow \quad v=\pm 20.7\,\rm m/s \end{gather*} As you can see, we obtained a speed with two different signs. Here, the speed with a positive sign indicates that the baseball at that desired height is being moved upward, while the negative sign hints to us that the baseball is at that height when it is moving down. 

Thus, we arrive at the fact that the baseball reaches that height twice, once when it is going up and once when it is going down.

(b) In the previous part, we found out that we were at that height twice. Thus, there are two corresponding times for this situation. Applying the equation $y-y_0=-\frac 12 gt^2+v_0t$ and solving for the required time $t$, we get \begin{gather*} 10-0=-\frac 12 (10)t^2+25t \\\\ \Rightarrow \quad 4.9t^2-25t+10=0 \end{gather*} The quadratic equation above has two solutions as below \begin{gather*} t_1=0.44\,\rm s \quad , \quad t_2=4.66\,\rm s \end{gather*} the first time, $t_1$ is when the ball is going up, and the second one corresponds to when it is moving down. 

Note : The quadratic equation of $at^2+bt+c=0$ has the following solution formula \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Problem (9): A helicopter is ascending vertically upward with a constant speed of $6.5\,\rm m/s$ and carrying a package of $2\,\rm kg$. When it reaches a height of $150\,\rm m$ above the surface, it drops the package. How much time does it take for the package to hit the surface?

Solution : Take up the positive direction, as always. At the moment that the package is released, it has the speed of its carrier. 

There is a subtle point in this case. Since the object is moving down, in the opposite direction of our positive direction, a negative must accompany it. Therefore, the correct input for the initial speed is $v_0=-6.5\,\rm m/s$.

Applying the equation $y-y_0=-\frac 12 gt^2+v_0t$, we will have \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ -150-0=-\frac 12 (10)t^2+(-6.5)t \\\\ \Rightarrow 5t^2+6.5t-150=0 \end{gather*} Using a graphing calculator or the formula in the previous question, we find that it takes about $4.9\,\rm s$ for the package to reach the ground. 

Note that in the above, we inserted a negative for the vertical height as $y=-150\,\rm m$, since the package hits the ground $150\,\rm m$ below our chosen origin. 

Problem (10): A stone is thrown vertically upward from a building $15\,{\rm m}$ high with an initial velocity of $10\,{\rm m/s}$. What is the stone's velocity just before hitting the ground?

Solution : In all freely falling practice problems, the important note is choosing the origin. Usually, the throwing (releasing or dropping) point is the best choice. In this case, the hitting point is below the origin, so its vertical displacement ($y$) is negative.

Applying the time-independent free fall kinematic equation, we have \begin{align*}v_f^{2}-v_i^{2}&=2\,(-g)\Delta y\\v_f^{2}-(10)^{2}&=2\,(-10)(-15) \\ \Rightarrow v_f&=\pm 20\,{\rm m/s}\end{align*} Since velocity is a vector quantity and just before striking the ground its direction is vertically downward, the negative value must be chosen, i.e., $v_f=-20\,{\rm m/s}$. 

Problem (11): A bullet is fired vertically downward with an initial speed of $15\,{\rm m/s}$ from the top of a tower of $20\,{\rm m}$ high. What is its velocity at the instant of striking the ground? 

Solution : Let the origin be the firing point. Using the below kinematic equation, we have \begin{align*}v_f^{2}-v_i^{2}&=-2g(y-y_0) \\\\ v_f^{2}-0&=-2(10)(-20)\\\Rightarrow v_f&=\pm 20\,{\rm m/s}\end{align*} Since the direction of velocity at the moment of striking the ground is downward, we must choose the negative correct sign, so we have $v_f=-20\,\rm m/s$. 

Problem (12): There is a well with a depth of $34\,{\rm m}$. A person drops a stone vertically into it with an initial velocity of $7\,{\rm m/s}$. What is the time interval between dropping the stone and hearing its impact sound? (Assume $g=10\,{\rm m/s^2}$ and the speed of the sound in the air is $340\,{\rm m/s}$).

Solution : This motion has two parts. One is descending into the well, which is a constant acceleration motion, and the other is ascending to the sound of impact, which is a uniform motion with constant speed. 

For the first part, use the kinematic equation $y-y_0=-\frac 12 gt^{2}+v_0 t$ to find the falling time as \begin{gather*}y-y_0=-\frac 12 gt^{2}+v_0 t \\\\ -34-0=-\frac 12 (10)t^{2}+(-7)t \\\\ \Rightarrow 5t^{2}+7t-34=0 \end{gather*} Since initial velocity is a vector and its direction is initially downward, a negative is included in front of it. The above quadratic equation has two solutions: $t_1=2\,{\rm s}$ and $t_2=-3.4\,{\rm s}$. Obviously, a negative value for time is not accepted because time in all kinematic questions must be a positive quantity.

The second part is a uniform motion as the speed of sound is constant, so we can calculate its rising time using the definition of average velocity: \begin{align*} t&=\frac{\Delta y}{v} \\\\ &=\frac{34}{340}=0.1\,{\rm s}\end{align*} Thus, the total time is obtained as $t_T=2+0.1=2.1\,{\rm s}$. 

Problem (13): A stone is dropped vertically downward from the top of a building with a height of $h$. What is its speed at the height of $\frac h2$?

Solution : Let the dropping point be the origin. Thus, in the kinematic equations, the vertical displacement must be negative, i.e., $y-y_0=-\frac h2$. Use the below equation to find the speed at the desired level \begin{align*}v_2^2-v_1^2&=-2g(y-y_0) \\\\ v_2^2 -0&=-2g(-\frac h2) \\\\ \Rightarrow v_2&=\sqrt{gh}\end{align*}

Problem (14): You throw a baseball straight up in the air and catch it 3.4 seconds later at the same place at which you threw it. How high did it go? (DIFFICULT)

Solution : First of all, choose a coordinate system with the upward direction as positive. In the absence of air resistance, when you throw an object upward, the time it takes to rise is equal to the time it takes to fall. In other words, the time of reaching the highest point is half the total time the object is in the air \[t_{rise}=t_{fall}=\frac 12 t_{tot}\] Other than the total time of flight, no other relevant information is given. Substitute the time taken to reach the highest point into $v=v_0-gt$ and solve for the initial velocity $v_0$ as below \begin{gather*} v_{top}=v_0-gt_{rise} \\\\ 0=v_0-(9.8)(1.7) \\\\ \Rightarrow v_0=16.6\,\rm m/s\end{gather*} Note that at the highest point $v_{top}=0$. 

Now, apply the equation $v^2-v_0^2=-2g\Delta y$ between the initial point and the highest point so that $\Delta y=h$ and solve for $h$. \begin{gather*} v^2_{top}-v_0^2=-2gh \\\\ (0)^2-(16.6)^2=-2(9.8)h \\\\ \Rightarrow \boxed{h=14.0\,\rm m} \end{gather*} Therefore, the ball rises as high as $14.0\,\rm m$.

Problem (15): A bullet is fired vertically upward from a height of $90\,{\rm m}$ and after $10\,{\rm s}$ reaches the ground. After $2\,{\rm s}$ from the throwing point, the bullet is how far away from the surface? ($g=9.8\,{\rm m/s^2}$)

Solution : In a free-fall problem, the vertical position of an object in an instant of time is given by the kinematic equation $y=-\frac 12 gt^{2}+v_0 t+y_0$. Let the firing point be the origin, so $y_0=0$. To complete the equation above, you must have an initial speed. 

To find the initial speed, apply the kinematic formula $y=-\frac 12 gt^{2}+v_0 t+y_0$ between the origin and striking point. \begin{gather*}y-y_0=-\frac 12 gt^{2}+v_0 t \\\\ -90-0=-\frac 12 (9.8)(10)^{2}+v_0(10) \\\\ \Rightarrow \boxed{v_0=40\,\rm m/s}\end{gather*} Notice that the striking point is $-90\,{\rm m}$ below the origin that's why the negative is entered for the vertical displacement $y$.  

Having the initial velocity, substitute it into the above equation again but with time $t=2\,{\rm s}$ to find the position of the bullet after $2\,{\rm s}$ with respect to the firing point \begin{align*}y-y_0&=-\frac 12 gt^{2}+v_0 t\\\\ &=-\frac 12\,(9.8)(2)^{2}+(40)(2) \\\\ \Rightarrow y&=\boxed{60.4\,\rm m} \end{align*}

Problem (16): A ball is thrown vertically upward with an initial velocity of $18\,\rm m/s$. How many seconds after throwing, the ball's speed is $9\,\rm m/s$ downward?

Solution : In all kinematic equations, $x,y,v,v_0$, and $a$ are vectors, so their signs matter. A speed of $9\,{\rm m/s}$ downward means a velocity of $-9\,{\rm m/s}$. Downward or upward indicates the direction of velocity. 

Take up as the positive $y$ direction. Now use the equation $v=v_0-gt$ to find the velocity at any later time.\begin{gather*}v=v_0-gt\\-9=+18-(10)t\\ \Rightarrow \quad \boxed{t=2.7\,\rm s}\end{gather*}

Problem (17): An object is thrown vertically upward in the air from a $100\,{\rm m}$ height with an initial velocity of $v_0$. After $5\,{\rm s}$, it reaches the ground. Determine the magnitude and direction of the initial velocity.

Solution : Let the origin be the throwing point (so $y_0=0$) and the upward direction positive. Substitute the given total time into the vertical displacement kinematic equation $y-y_0=-\frac 12 gt^2+v_0t$ with $y-y_0=-100\,{\rm m}$ (since the impact point is below the origin). \begin{gather*} y-y_0 =-\frac 12 gt^2+v_0t \\\\ -100=-\frac 12\,(10)(5)^2+v_0 (5) \\\\ \Rightarrow \boxed{v_0=-5\,\rm m/s}\end{gather*} the minus sign indicates the initial velocity is downward with the magnitude (speed) of $5\,\rm m/s$.

Problem (18): From a height of $15\,{\rm m}$, a ball is kicked vertically up into the air with an initial speed of $v_0$. It reaches the highest point of its path with an elevation of $20\,{\rm m}$ from the surface. Find the initial velocity $v_0$. 

Solution : The highest point is $5\,{\rm m}$ above the kicking point. Apply the time-independent kinematic equation below to find the initial velocity \begin{gather*}v^2-v_0^2=-2g(y-y_0) \\\\ 0-v_0^2=-2(10)(5) \\\\ \Rightarrow v_0=\pm 10\,{\rm m/s}\end{gather*} Because the ball kicked upward so we must choose the plus sign, i.e. $v_0=+10\,{\rm m/s}$. In the above, we used the fact that in all freely falling problems, at the highest point (apex) the velocity is zero ($v=0$).

Recall that projectiles are a particular type of freely falling motion with a launch angle of $\theta=90$ with its own formulas.

Problem (19): From the bottom of a $25\,{\rm m}$-depth well, a stone is thrown vertically upward with an initial speed of $30\,{\rm m/s}$.  (a) How high does the stone rise out of the well?  (b) Before the stone returns into the well, how many seconds are outside the well?

Solution :  (a) Let the bottom of the well be the origin, so $y_0=0$. First, we find how much distance the ball rises. Recall that the highest point is where $v=0$, so we have \begin{gather*}v^2-v_0^2=-2g(y-y_0) \\\\ 0-(30)^2=-2(10)(y-0) \\\\ \Rightarrow \boxed{y=45\,\rm m}\end{gather*} Of this height, $25\,{\rm m}$ is for the well's depth, so the stone is $20\,{\rm m}$ outside of the well. 

(b) We want to examine the duration between exiting and reentering the stone into the well. During this time interval, the ball returns to its initial position, so its displacement vector is zero, i.e., $\Delta y=y-y_0=0$. If we want to use the equation, $y-y_0=-\frac 12 gt^2+v_0t$, the speed of the stone is required exactly when it leaves the well. 

The speed at the bottom of the well is known. Thus, apply the equation $v^2-v_0^2=-2g\Delta y$ to find the speed just before it leaves the well. \begin{gather*}v^2-v_0^2=-2g\Delta y \\\\ v^2-(30)^{2}=-2(10)(25) \\\\ \Rightarrow v=+20\,{\rm m/s}\end{gather*} This speed can be used as the initial speed for the part where the stone is outside the well. Hence, the total time the stone is out of the well is obtained as below \begin{gather*} \Delta y=-\frac 12 gt^{2}+v_0 t\\ 0=-\frac 12 (10)t^{2}+(20)(2) \end{gather*} Solving for $t$, one can obtain the required time as $t=4\,{\rm s}$.

Problem (20): From the top of a $20-{\rm m}$-high tower, a small ball is thrown vertically upward. If $4\,{\rm s}$ after throwing, it hit the ground. How many seconds before striking the surface does the ball again meet the original throwing point? (Air resistance is neglected and $g=10\,{\rm m/s^2}$). 

Solution : Let the origin be the throwing point. The ball strikes the ground $20\,\rm m$ below our chosen origin, so its total displacement between initial position $y_i=0$ and final position is $\Delta y=y_f-y_i=-20\,\rm m$. The total time in which the ball is in the air is also $4\,{\rm s}$. With these known values, one can find the initial velocity as \begin{gather*}\Delta y=-\frac 12 gt^{2}+v_0t \\\\ -25=-\frac 12 (10)(4)^{2}+v_0(4) \\\\ \Rightarrow \boxed{v_0=15\,\rm m/s} \end{gather*} When the ball returns to its initial position, its total displacement is zero, i.e., $\Delta y=0$ so we can use the following kinematic equation to find the total time it takes the ball to return to the starting point \begin{gather*}\Delta y=-\frac 12 gt^{2}+v_0t \\\\ 0=-\frac 12 (10)t^{2}+(15)t \end{gather*} Rearranging and solving for $t$, we get $t=3\,{\rm s}$.

Problem (21): A rock is thrown vertically upward in the air. It reaches the height of $40\,{\rm m}$ from the surface at times $t_1=2\,{\rm s}$ and $t_2$. Find $t_2$ and determine the greatest height reached by the rock (neglect air resistance and assume $g=10\,{\rm m/s^2}$).

Solution : Let the throwing point (surface of the ground) be the origin. Between our chosen origin and the point with known values $h=4\,{\rm m}$, $t=2\,{\rm s}$ one can write down the kinematic equation $\Delta y=-\frac 12 gt^{2}+v_0\,t$ to find the initial velocity as \begin{gather*} \Delta y=-\frac 12 gt^{2}+v_0t \\\\ 40=-\frac 12 (10)(2)^2 +v_0(2) \\\\ \Rightarrow v_0=30\,{\rm m/s}\end{gather*} Now we are going to find the times when the rock reaches the height $40\,{\rm m}$ (Recall that when an object is thrown upward, it passes through every point twice). Applying the same equation above, we get \begin{gather*} \Delta y=-\frac 12 gt^{2}+v_0t \\\\ 40=-\frac 12\,(10)t^2+30t \\\\ \Rightarrow \boxed{5t^2-30t+40=0} \end{gather*} Rearranging and solving for $t$ using quadratic equation formula, two times are obtained i.e. $t_1=2\,{\rm s}$ and $t_2=4\,{\rm s}$. Thus, again after $4\,\rm s$ the ball again is at the same height of $40\,\rm m$ from the surface. 

The greatest height is where the vertical velocity becomes zero so we have \begin{gather*}v^2-v_0^2 =-2g\Delta y \\\\ 0-(30)^2=-2(10)\Delta y\\\\ \Rightarrow \boxed{\Delta y=45\,\rm m}\end{gather*} Thus, the highest point where the rock can reach is located at $H=45\,{\rm m}$ above the ground.

Problem (22): A ball is launched with an initial velocity of $30\,{\rm m/s}$ straight upward. How long will it take the ball to reach $20\,{\rm m}$ below the highest point for the first time? (neglect air resistance and assume $g=10\,{\rm m/s^2}$).

Solution : Between the origin (surface level) and the highest point ($v=0$) apply the time-independent kinematic equation below to find the greatest height $H$ where the ball reaches.\begin{gather*}v^2-v_0^2=-2g\Delta y \\\\ 0-(30)^2=-2(10)H \\\\ \Rightarrow \boxed{H=45\,\rm m}\end{gather*} This is the maximum height that the ball can reach. The $20\,{\rm m}$ below this maximum height $H$ has a height of $h=45-20=25\,{\rm m}$. Now use the vertical displacement kinematic equation between the throwing point and the desired position to find the required time taken. \begin{gather*} \Delta y=-\frac 12 gt^2+v_0t \\\\ 25=-\frac 12(10)t^2+30(t)\end{gather*} Solving for $t$ (using quadratic formula), we get $t_1=1\,{\rm s}$ and $t_2=5\,{\rm s}$ one for up way and the second for down way.  

Problem (23): A stone is launched directly upward from the surface level with an initial velocity of $20\,{\rm m/s}$. How many seconds after launch is the stone's velocity $5\,{\rm m/s}$ downward?

Solution : Let the origin be at the surface level and take the positive direction up. Therefore, we have initial velocity $v_0=+20\,{\rm m/s}$ and final velocity $v=-5\,{\rm m/s}$. Use the velocity kinematic equation $v=v_0-gt$ to find the desired time as below \begin{gather*}v=v_0-gt \\-5=+20-10\times t \\ \Rightarrow \boxed{t=2.5\,\rm s}\end{gather*}

Problem (24): From a $25-{\rm m}$ building, a ball is thrown vertically upward at an initial velocity of $20\,{\rm m/s}$. How long will it take the ball to hit the ground?

Solution : Origin is considered to be at the throwing point, so $y_0=0$. Apply the position kinematic equation below to find the desired time \begin{gather*} y-y_0=-\frac 12 gt^2+v_0 t \\\\ -25=-\frac 12(10)t^2+20t \\\\ \Rightarrow 5t^2-20t-25=0 \end{gather*} Rearranging and converting it into the standard form of a quadratic equation $at^2+bt+c=0$, its solutions are obtained as \begin{align*}t_{1,2}&=\frac{-b\pm \sqrt{b^2-4\,ac}}{2a} \\\\ &=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-5)}}{2(1)}\\\\ &=-1 \, \text{and} \, 5 \end{align*} Therefore, the time needed for the ball to hit the ground is $5\,{\rm s}$.

Problem (25): From the top of a building with a height of $60\,{\rm m}$, a rock is thrown directly upward at an initial velocity of $20\,{\rm m/s}$. What is the rock's velocity at the instant of hitting the ground? 

Solution : Apply the time-independent kinematic equation as \begin{gather*} v^2-v_0^2 =-2g(y-y_0) \\\\ v^2-(20)^2 =-2(10)(-60) \\\\ v^2 =1600\\\\ \Rightarrow \quad \boxed{v=\pm 40\,\rm m/s} \end{gather*} Therefore, the rock's velocity when it hit the ground is $v=-40\,{\rm m/s}$.

In this article, you learned how to solve free fall problems step-by-step using simple kinematic equations. 

Author : Dr. Ali Nemati Published : 8/11/2022  

© 2015 All rights reserved. by Physexams.com

An hour and thirty-one minutes after launch, my pressure altimeter halts at 103,300 feet. At ground control the radar altimeters also have stopped on readings of 102,800 feet, the figure that we later agree upon as the more reliable. It is 7 o'clock in the morning, and I have reached float altitude. At zero count I step into space. No wind whistles or billows my clothing. I have absolutely no sensation of the increasing speed with which I fall. Though my stabilization chute opens at 96,000 feet, I accelerate for 6,000 feet more before hitting a peak of 614 miles an hour, nine-tenths the speed of sound at my altitude. An Air Force camera on the gondola took this photograph when the cotton clouds still lay 80,000 feet below. At 21,000 feet they rushed up so chillingly that I had to remind myself they were vapor and not solid. Joseph Kittinger, 1960

• just before it hit the floor on the way down
• just after it left the floor on the way up
• on the way down
• while it is contact with the floor
• on the way up
• the time the diver was in the air
• the maximum height to which she ascended
• her velocity on impact with the water
• displacement-time
• velocity-time
• acceleration-time
• at it highest point
• Whole body reaction time is about three-tenths of a second. Would you have enough time to throw yourself clear of a brick falling from a point 3 m directly overhead if you saw it the moment it came loose?
• straight up
• straight down
• Determine the minimum time needed to reach to reach this velocity for a skydiver starting from rest.
• Determine the minimum displacement needed to reach to reach this velocity for a skydiver starting from rest.
• Why are these values minimums?
• At what altitude might one of these skydivers break through the sound barrier? Assume that the acceleration due to gravity is 9.70 m/s 2 , the speed of sound is 300 m/s, and that air resistance is negligible.
• Captain Kittinger believed that air resistance was negligible down to about 27.5 km (90,000 feet). Assuming a continued acceleration of 9.72 m/s 2  after exceeding the speed of sound, determine a possible maximum speed during such a jump.
• 10 m above the point at which it was released?
• 20 m above the point at which it was released?
• 30 m above the point at which it was released?
• Determine the initial velocity of the stone.
• At what time does this occur?
• At what height does this occur?
• When do the two acrobats pass each other?
• At what height above the trampoline are they?
• What are their respective velocities?

During a simple drop experiment, the capsule is pulled up to a height of 120 meters to the top of the drop tube and then released. After 4.74 seconds the experiment has landed safely in the deceleration unit filled with polystyrene pellets. Before the experiment, 18 high-performance pumps make sure that the drop tube is almost free of air containing only one ten thousandth of the normal air pressure. Due to the vacuum, the air drag is so low that the Bremen Drop Tower can provide one of the best quality of microgravity — in some aspects even better than on the International Space Station (ISS). Therefore, ZARM's drop tower facility is a very economic and easily accessible alternative to doing research in space. ZARM, 2020

• Over what distance does the capsule fall freely? (This is not the same as the height of the drop tube or the height of the tower.)
• What is the impact speed of a capsule?
• What is the length of the deceleration unit?
• What is the average acceleration (magnitude and direction) of a capsule in the deceleration unit?

The pneumatically driven system takes 0.25 seconds to accelerate the experiment capsule to a speed of 168 kilometers per hour. The exact force of acceleration is being calculated for each individual experiment in order to throw the drop capsule as close as possible to the top of the drop tube and thus maximize the duration of the flight. After a couple of seconds the deceleration unit has already been moved into place again in order to catch the capsule on its way down. ZARM, 2020

• What is the launch speed of the capsule in meters per second?
• What is the average acceleration (magnitude and direction) of the capsule while it is being pushed up in the catapult?
• Over what distance does the catapult push a capsule?
• For how long is a capsule in free fall after a launch? Or in other words, how long does it take a capsule to return to the height from which it was launched?
• How high does the capsule rise after being lauched from the catapult? (This is not the same as the height of the drop tube or the height of the tower or the answer to any previous part of this question.)
• How far did Lois Lane fall before Superman caught her?
• For how long was Lois Lane falling?
• How long did it take Superman and Lois Lane to stop?
• How far did they travel while stopping?
• How tall is the Daily Planet Building?
• The diagram to the right shows a US one dollar bill held vertically above a person's hand with their finger and thumb arranged like they are ready to pinch something. Fingertip reaction time is about 200 ms for an average person. Show that it is unlikely that someone would be able to catch a US dollar bill in a situation like this if it was released without warning.

investigative

• You can determine a person's reaction time using a centimeter ruler. Find several volunteers and have them hold their open hand out so that you can drop a ruler between their thumb and fingers. Suspend the ruler vertically with the zero mark of the ruler at the same level as the top of your volunteer's open hand. Tell your volunteers to grab the ruler the instant you drop it. Release it without warning and record the position on the ruler where they grabbed it. (Take the reading from the top of their fingers.) Repeat this test a few times for each volunteer to obtain an average value. Record your results along with your volunteer's ages (or another demographic variable that you think may be relevant). Determine the relation, if any, between age and reaction time in your sample. (A list of demographic variables to test might include such things as gender, place of birth, hours spent playing video games, hours spent watching TV, number of siblings, whatever.)
• using the time a pop fly is in the air or
• using the speed of a batted ball.
• How long was Lois Lane in apparent free fall?
• What speed would she have when Superman caught her?
• How far would she have fallen in this time?
• Describe the physical realism of this part of the scene.
• How long did it take Superman to stop Lois Lane? (You may need to count frames. There are 25 frames per second in this video.)
• Calculate Lois Lane's deceleration during the catch assuming she was traveling at the speed you calculated in part b.
• Determine the distance needed to slow Lois Lane to a stop in the time given in part d.
• Describe the likelihood of surviving a rescue attempt like this one.

A second opinion

• Watch this video clip from The Big Bang Theory , Season 1, Episode 2: The Big Bran Hypothesis (1 October 2007). Was Sheldon right about Lois Lane's chances of surviving a fall like the one in the movie?

youtu.be/hTujOtTqSsQ

• How many seconds did it take Mr. Brown to fall from his highest position until he just touched the floor?
• What speed did he have on impact?
• What height did he fall?
• How long did it take him to stop after he touched the floor?
• What was his average deceleration?

• When did the blob jumper reach the highest point of the jump? How long did it take him to hit the lake surface after reaching the highest point of the jump? Use these numbers to answer the remaining parts of this problem.
• How did his velocity on take off compare to his velocity on landing? Determine both velocities (their magnitudes and directions).
• How high did he rise? Confirm or refute the claim that the jumper hit 17 m.
• For how long was the rock falling? (The playback speed is 14.81 frames per second.)
• How far did the rock fall?
• What speed did the rock have when it struck the water?
• How long did it take the rock to fall half the distance you calculated in part b?
• How long did it take the rock to reach half the speed you calculated in part c?
• Determine the deceleration of the rock after it struck the water.
• Determine the time it took the rock to stop.

• the height the cat fell
• the speed it had when it hit the grass

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Application: Falling Objects (1D)

Application: projectile motion (2d), forces & newton's laws of motion, potential energy, thermal energy.

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Free Fall (Physics): Definition, Formula, Problems & Solutions (w/ Examples)

​ Free fall ​ refers to situations in physics where the only force acting on an object is gravity.

The simplest examples occur when objects fall from a given height above the surface of the Earth straight downward – a one-dimensional problem. If the object is tossed upward or forcefully thrown straight downward, the example is still one-dimensional, but with a twist.

Projectile motion is a classic category of free-fall problems. In reality, of course, these events unfold in the three-dimensional world, but for introductory physics purposes, they are treated on paper (or on your screen) as two-dimensional: ​ x ​ for right and left (with right being positive), and ​ y ​ for up and down (with up being positive).

Free-fall examples therefore often have negative values for y-displacement.

It is perhaps counterintuitive that some free-fall problems qualify as such.

Keep in mind that the only criterion is that the only force acting on the object is gravity (usually Earth's gravity). Even if an object is launched into the sky with colossal initial force, at the moment the object is released and thereafter, the only force acting on it is gravity and it is now a projectile.

• Often, high-school and many college physics problems neglect air resistance, though this always has at least a slight effect in reality; the exception is an event that unfolds in a vacuum. This is discussed in detail later.

The Unique Contribution of Gravity

A unique an interesting property of the acceleration due to gravity is that it is the same for all masses.

This was far from self-evident until the days of Galileo Galilei (1564-1642). That's because in reality gravity is not the only force acting as an object falls, and the effects of air resistance tend to cause lighter objects to accelerate more slowly – something we've all noticed when comparing the fall rate of a rock and a feather.

Galileo conducted ingenious experiments at the "leaning" Tower of Pisa, proving by dropping masses of different weights from the high top of the tower that gravitational acceleration is independent of mass.

Solving Free-Fall Problems

Usually, you are looking to determine initial velocity (v 0y ), final velocity (v y ) or how far something has fallen (y − y 0 ). Although Earth's gravitational acceleration is a constant 9.8 m/s 2 , elsewhere (such as on the moon) the constant acceleration experienced by an object in free fall has a different value.

For free fall in one dimension (for example, an apple falling straight down from a tree), use the kinematic equations in the ​ Kinematic Equations for Free-Falling Objects ​ section. For a projectile-motion problem in two dimensions, use the kinematic equations in the section ​ Projectile Motion and Coordinate Systems ​.

• You can also use the conservation of energy principle, which states that ​ the loss of potential energy (PE) ​ during the fall ​ equals the gain in kinetic energy (KE): ​ –mg(y − y 0 ) = (1/2)mv y 2 . 

Kinematic Equations for Free-Falling Objects

All of the foregoing can be reduced for present purposes to the following three equations. These are tailored for free fall, so that the "y" subscripts can be omitted. Assume that acceleration, per physics convention, equals −g (with the positive direction therefore upward).

• Note that v 0 and y 0  are initial values in any problem, not variables.

​ Example 1: ​ A strange birdlike animal is hovering in the air 10 m directly over your head, daring you to hit it with the rotten tomato you're holding. With what minimum initial velocity v 0 would you have to throw the tomato straight up in order to ensure that it reaches its squawking target?

What's happening physically is that the ball is coming to a stop owing to the force of gravity just as it reaches the required height, so here, v y = v = 0.

First, list your known quantities: ​ v = ​ 0​ , g = ​ –9.8 m/s2​ , y − y 0 = ​ 10 m

Thus you can use the third of the equations above to solve:

This is about 31 miles an hour.

Projectile Motion and Coordinate Systems

Projectile motion involves the motion of an object in (usually) two dimensions under the force of gravity. The behavior of the object in the x-direction and in the y-direction can be described separately in assembling the greater picture of the particle's motion. This means that "g" appears in most of the equations required to solve all projectile-motion problems, not merely those involving free fall.

The kinematic equations needed to solve basic projectile motion problems, which omit air resistance:

​ Example 2: ​ A daredevil decides to try to drive his "rocket car" across the gap between adjacent building rooftops. These are separated by 100 horizontal meters, and the roof of the "take-off" building is 30 m higher than the second (this almost 100 feet, or perhaps 8 to 10 "floors," i.e., levels).

Neglecting air resistance, how fast will he need to be going as he leaves the first rooftop to assure just reaching the second rooftop? Assume his vertical velocity is zero at the instant the car takes off.

Again, list your known quantities: (x – x 0 ) = 100m, (y – y 0 ) = –30m, v 0y = 0, g = –9.8 m/s 2 .

Here, you take advantage of the fact that horizontal motion and vertical motion can be assessed independently. How long the car will take to free-fall (for purposes of y-motion) 30 m? The answer is given by y – y 0 = v 0y t − (1/2)gt 2.

Filling in the known quantities and solving for t:

Now plug this value into x = x 0 + v 0x t :

​ v 0x = 40.4 m/s (about 90 miles per hour). ​

This is perhaps possible, depending on the size of the roof, but all in all not a good idea outside of action-hero movies.

Hitting it out of the Park... Far Out

Air resistance plays a major, under-appreciated role in everyday events even when free fall is only part of the physical story. In 2018, a professional baseball player named Giancarlo Stanton hit a pitched ball hard enough to blast it away from home plate at a record 121.7 miles per hour.

The equation for the maximum horizontal distance a launched projectile can attain, or ​ range equation ​ (see Resources), is:

Based on this, if Stanton had hit the ball at the theoretical ideal angle of 45 degrees (where sin 2θ is at its maximum value of 1), the ball would have traveled 978 feet! In reality, home runs almost never reach even 500 feet. Part if this is because a launch angle of 45 degrees for a batter is not ideal, as the pitch is coming in almost horizontally. But much of the difference is owed to the velocity-dampening effects of air resistance.

Air Resistance: Anything But "Negligible"

Free-fall physics problems aimed at less advanced students assume the absence of air resistance because this factor would introduce another force that can slow or decelerate objects and would need to be mathematically accounted for. This is a task best reserved for advanced courses, but it bears discussion here nonetheless.

In the real world, the Earth's atmosphere provides some resistance to an object in free fall. Particles in the air collide with the falling object, which results in transforming some of its kinetic energy into thermal energy. Since energy is conserved in general, this results in "less motion" or a more slowly increasing downward velocity.

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About the Author

Kevin Beck holds a bachelor's degree in physics with minors in math and chemistry from the University of Vermont. Formerly with ScienceBlogs.com and the editor of "Run Strong," he has written for Runner's World, Men's Fitness, Competitor, and a variety of other publications. More about Kevin and links to his professional work can be found at www.kemibe.com.

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• Physics Formulas

Free Fall Formula

Freefall as the term says, is a body falling freely because of the gravitational pull of our earth.

Imagine a body with velocity (v) is falling freely from a height (h) for time (t) seconds because of gravity (g).

Free Fall Formulas  are articulated as follows:

h = (1/2) gt 2

Freefall Related Solved Examples

Underneath are given questions on free fall which may be useful for you.

Problem 1:  Calculate  the  body  height  if  it  has  a  mass  of  2  kg  and  after  7 seconds  it  reaches  the  ground?

Given: Height h =? Time t = 7s We all are acquainted with the fact that free fall is independent of mass.

Hence, it is given as

h = 0.5 × 9.8 × (7) 2

h = 240.1 m

Problem  2: The cotton falls after 3 s and iron falls after 5 s. Which is moving with higher velocity? Answer:

The Velocity in free fall is autonomous of mass.

V (Velocity of iron) = gt = 9.8 m/s 2  × 5s = 49 m/s

V (Velocity of cotton) = gt = 9.8 m/s 2  × 3s = 29.4 m/s.

The Velocity of iron is more than cotton.

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Free Falling Body

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One of the most common sorts of problems that a beginning physics student will encounter is to analyze the motion of a free-falling body. It's helpful to look at the various ways these sorts of problems can be approached.

The following problem was presented on our long-gone Physics Forum by a person with the somewhat unsettling pseudonym "c4iscool":

A 10kg block being held at rest above the ground is released. The block begins to fall under only the effect of gravity. At the instant that the block is 2.0 meters above the ground, the speed of the block is 2.5 meters per second. At what height was the block released?

Begin by defining your variables:

• y 0 - initial height, unknown (what we're trying to solve for)
• v 0 = 0 (initial velocity is 0 since we know it begins at rest)
• y = 2.0 m/s
• v = 2.5 m/s (velocity at 2.0 meters above ground)
• m = 10 kg
• g = 9.8 m/s 2 (acceleration due to gravity)

Looking at the variables, we see a couple of things that we could do. We can use conservation of energy or we could apply one-dimensional kinematics .

Method One: Conservation of Energy

This motion exhibits conservation of energy, so you can approach the problem that way. To do this, we'll have to be familiar with three other variables:

• U = mgy ( gravitational potential energy )
• K = 0.5 mv 2 ( kinetic energy )
• E = K + U (total classical energy)

We can then apply this information to get the total energy when the block is released and the total energy at the 2.0-meter above-the-ground point. Since the initial velocity is 0, there is no kinetic energy there, as the equation shows

E 0 = K 0 + U 0 = 0 + mgy 0 = mgy 0 E = K + U = 0.5 mv 2 + mgy by setting them equal to each other, we get: mgy 0 = 0.5 mv 2 + mgy and by isolating y 0 (i.e. dividing everything by mg ) we get: y 0 = 0.5 v 2 / g + y

Notice that the equation we get for y 0 doesn't include mass at all. It doesn't matter if the block of wood weighs 10 kg or 1,000,000 kg, we will get the same answer to this problem.

Now we take the last equation and just plug our values in for the variables to get the solution:

y 0 = 0.5 * (2.5 m/s) 2 / (9.8 m/s 2 ) + 2.0 m = 2.3 m

This is an approximate solution since we are only using two significant figures in this problem.

Method Two: One-Dimensional Kinematics

Looking over the variables we know and the kinematics equation for a one-dimensional situation, one thing to notice is that we have no knowledge of the time involved in the drop. So we have to have an equation without time. Fortunately, we have one (although I'll replace the x with y since we're dealing with vertical motion and a with g since our acceleration is gravity):

v 2 = v 0 2 + 2 g ( x - x 0 )

First, we know that v 0 = 0. Second, we have to keep in mind our coordinate system (unlike the energy example). In this case, up is positive, so g is in the negative direction.

v 2 = 2 g ( y - y 0 ) v 2 / 2 g = y - y 0 y 0 = -0.5 v 2 / g + y

Notice that this is exactly the same equation that we ended up within the conservation of energy method. It looks different because one term is negative, but since g is now negative, those negatives will cancel and yield the exact same answer: 2.3 m.

Bonus Method: Deductive Reasoning

This won't give you the solution, but it will allow you to get a rough estimate of what to expect. More importantly, it allows you to answer the fundamental question that you should ask yourself when you get done with a physics problem:

Does my solution make sense?

The acceleration due to gravity is 9.8 m/s 2 . This means that after falling for 1 second, an object will be moving at 9.8 m/s.

In the above problem, the object is moving at only 2.5 m/s after having been dropped from rest. Therefore, when it reaches 2.0 m in height, we know that it hasn't fallen very fall at all.

Our solution for the drop height, 2.3 m, shows exactly this; it had fallen only 0.3 m. The calculated solution does make sense in this case.

• One-Dimensional Kinematics: Motion Along a Straight Line
• The Difference Between Terminal Velocity and Free Fall
• An Idealized Model in Physics
• Two-Dimensional Kinematics or Motion in a Plane
• Introduction to Newton's Laws of Motion
• What Is the Definition of Work in Physics?
• What Is a Thermodynamic Process?
• An Introduction to Density: Definition and Calculation
• Understanding Cosmology and Its Impact
• Five Great Problems in Theoretical Physics
• Doppler Effect in Light: Red & Blue Shift
• Fundamental Physical Constants
• Black Holes and Hawking Radiation
• Kinetic Molecular Theory of Gases
• What Is Moment of Inertia in Physics?
• Understanding the "Schrodinger's Cat" Thought Experiment

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Chapter 3: Motion Along a Straight Line

Back to chapter, free-falling bodies: introduction, previous video 3.10: kinematic equations: problem solving, next video 3.12: free-falling bodies: example.

When an object falls towards the ground and is only influenced by gravity, it is said to be in free-fall.

Consider a ball thrown upwards in the air, reaches a certain height before falling back due to Earth's gravitational force.

Here, the ball is undergoing changes in velocity during its fall, but the acceleration is constant.

Consider a paper and a stone both being dropped at the same time and from the same height. Which object will reach the ground first?

It is observed that the stone reaches the ground first, followed by paper. This is due to the air resistance, which is negligible for the heavy objects.

However, if the objects are dropped together from the same height in a closed vacuum box, they will reach the base of the box at the same time.

Therefore, if the air effects are ignored, all objects irrespective of their masses, fall with a constant acceleration, known as acceleration due to gravity, which has an approximate value of 9.8 m/s 2 .

All objects, neglecting air resistance, fall with the same acceleration towards the Earth's center due to the force exerted by the Earth's gravity. This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavier ones. People believed that a heavier object had a greater acceleration when falling until Galileo Galilei (1564–1642) proved otherwise. We now know this is not the case.

The motion of an object under the influence of gravitational force is called a free-fall. Free-fall motion is not only associated with objects dropped from rest, but with all objects moving freely under gravity's influence. The acceleration of objects under a free-fall is constant, and is known as acceleration due to gravity, g . The magnitude of acceleration due to gravity is 9.8 m/s 2 . Because of the constant acceleration, kinematic equations of motion can be used to predict the dynamics of objects under free-fall, provided that the air effects are neglected.

In the real world, however, air resistance is always present, and it opposes the motion of objects. The momentum required to oppose the motion of heavier objects is larger compared to lighter objects. Therefore, heavier objects fall faster towards the ground during free-fall in the presence of air resistance.

This text is adapted from Openstax, University Physics Volume 1, Section 3.5: Free-fall .

Free fall motion – problems and solutions

Wanted : The final velocity of the stone when it hits the ground (v t )

v t = √900 = 30 m/s 2

The equation of the free fall motion :

Wanted : Time interval (t)

Acceleration due to gravity = g

(2)(2) = g t 1 2

1 = 1/2 g t 2 2

5. An object dropped from a height of h above the ground. The final velocity when object hits the ground is 10 m/s. What is the time interval to reach ½ h above the ground. Acceleration due to gravity is 10 m/s 2 .

The height of h :

The height of 1/2 h = 1/2 (5 meters) = 2.5 meters. The time interval needed to reach 2.5 meters above the ground :

Figure 2 = vertical motion = Deceleration

Known: time interval (t) and acceleration due to gravity (g), wanted: height (h) so use the equation of free fall motion: h = ½ g t 2

Acceleration due to gravity (g) = 10 m/s 2

The height of the fruit above the ground after 1 second :

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