# Falling Body Problems in Physics

These are my notes on falling body problems in physics.

The kinematic equations of motion for constant acceleration can be applied to a large collection of problems known as falling body problems. These problems are where the constant acceleration is the acceleration due to gravity on the surface of the Earth. I will list them here.

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\[v_{avg} = \frac{v+v_{0}}{2}\]

\[v = v_{0} + at\]

\[x - x_{0} = (½)(v + v_{0})t\]

\[x - x_{0} = v_{0}t + (½)at^{2}\]

\[v^{2} = v_{o}^{2} + 2a(x - x_{o})\]

**Example 1**

Consider a ball dropped from the top of a 40 m tall building. Calculate everything possible.

You will not be able to calculate everything at first.

The trick is to calculate what you can and those answers will lead you to the rest of the calculations.

We know:

displacement=40 meters

\[x=0\text{ } v_{o}=0\text{ } a=9.8m/s^{2}\]

Since most of the kinematic equations contain the time, this is usually one of the first things to calculate.

Use the equation that has all the variables included in it.

\[x-x_{0}=v^{2}_{0}t+(½)at^{2}\]

\[40m = (½)(9.8m/s^{2})t{2}\]

\[= 2.9s\]

Knowing the time, we can calculate the velocity using the second equation.

\[v = v_{0} + at\]

\[v = 0 + 9.8m/s^{2}(2.9)\]

\[= 28m/s\]

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**Example 2**

Now, using the original problem, let us add an initial velocity of 8.0m/s when the ball is thrown down instead of just dropped. Find the time for the ball to reach the ground and the velocity on impact.

First, we calculate the time for the ball to hit the ground as before.

\[x - x_{0} = v_{0}t + (½)at^{2}\]

\[40m = 8.0t + (½)(9.8t^{2}\]

\[40m = 8.0t + 4.9t^{2}\]

\[4.9t^{2} + 8.0t - 40m = 0\]

Use a quadratic formula.

\[t = \frac{-80\pm\sqrt{64-4(4.9)(-40)}}{2(4.9)}\]

\[t = 2.2s\]

The velocity at the ground level is:

\[v^{2} = (8.0m/s^{2})^{2} + 2(9.8m/s^{2})(40m) = 848m^{2}s^{2} = 29m/s\]

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**Example 3**

Using the first problem, now throw the ball up from the top of the building with a velocity of 8.0 m/s. Find the time for the ball to reach the ground and the velocity on impact.

Take x as the displacement as positive down and the velocity as negative up. It is important to remember that the sign of the velocity is opposite that of the displacement. It does not matter whether the velocity is negative and the distance down is positive, only that they are opposite. Calculate the time of flight.

\[40m = (-8.0m/s)t + (4.9m/s^{2})t^{2}\]

\[4.9t^{2} - 8.0t - 40 = 0\]

\[t = \frac{8.0\pm\sqrt{64-4(4.9)(-40)}}{9.8} = 3.8s\]

The positive time is the correct choice.

The velocity when the ball strikes the ground is:

\[v^{2} = (-8.0m/s^{2}) + 2(9.8m/s^{2})40m = 848m^{2}/s^{2} = 29m/s\]

Notice that whether the v term is a positive number or negative number, the result is the same. If the ball is thrown up with a certain velocity or down with the same velocity, the velocity at impact is the same. This is to be expected from the symmetry of the equations. If the ball is thrown up with a certain velocity, then on the way down it passes the same level with that same velocity.

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**Example 4**

For the situation in the above problem, calculate the maximum height above the top of the building and the time for the ball to reach maximum height.

The time for the ball to reach maximum height is from equation 2. Note that at maximum height, the velocity must be zero. Watch the signs closely. It does not matter how you choose the signs, but the acceleration has to be opposite the velocity.

\[v = v_{0} + at\]

\[0 = 8.0m/s - (9.8m/s^{2})\]

\[t = 0.82s\]

Because of the symmetry, it takes the ball the same amount of time to reach maximum height as it does for the ball to return to the original level.

Calculate the height above the top of the building from equation 5.

\[v^{2} = v_{o}^{2} + 2a(x - x_{o})\]

\[0^{2} = (8.0m/s^{2}) - 2(9.8m/s^{2})(x - x_{0}) = 3.3m\]

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**Example 5**

A bottle of champagne is dropped by a balloonist. The balloon is rising at a constant velocity of 3.0 m/s. It takes 8.0 s for the bottle of champagne to reach the ground. Find the height of the balloon when the bottle was dropped, the height of the balloon when the bottle reached the ground, and the velocity with which the bottle strikes the ground.

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There are several possibilities concerning the origin and direction of the coordinate system. Take the origin at the height of the balloon when the bottle is dropped and the position of the balloon at t=0. Take the displacement as positive down. The main reason for taking the displacement as positive down is that the acceleration is down and the initial velocity is up, making two positives and one negative. As time goes on, displacement, velocity, and acceleration will be positive.

First, write the equation for the height of the balloon starting from the time when the bottle is dropped.

\[x = (-3.0m/s)t + (4.9m/s^{2})t^{2} = (-3.0m/s)(8.0s) + (4.9m/s^{2})(64s^{2}) = 290m\]

The height of the balloon when the bottle was dropped plus the amount the balloon rose in the 8.0s it took the bottle to reach the ground:

\[290m + (3.0m/s)8.0s = 314m\]

The velocity on impact :

\[v = v_{0} + at = -3.0m/s + (9.8m/s^{2})(8.0s) = 75m/s\]

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**Example 6**

A parachutist descending at a constant rate of 2.0 m/s drops a smoke canister at a height of 300m. Find the time for the smoke canister to reach the ground and its velocity when it strikes the ground. Then find the time for the parachutist to reach the ground, the position of the parachutist when the smoke canister strikes the ground, and an expression for the distance between the smoke canister and the parachutist.

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The time for the smoke canister to reach the ground is from equation 4.

\[x = v_{0}t + (½)at^{2}\]

\[300m = (2.0m/s)t + (4.9m/s^{2})t^{2}\]

Without units, the equation is \(4.9t^{2} + 2.0t - 300 = 0\) with solutions.

\[t = \frac{-2\pm\sqrt{4-4(4.9)(-300)}}{2*4.9}\]

\[= \frac{-2\pm76.7}{9.8} = 7.6\]

The time for the canister to hit the ground is 7.6s.

The velocity when it strikes the ground is:

\[v^{2} = (2.0m/s)^{2} + 2(9.8m/s^{2})300m = 5884m^{2}/s^{2}= 77m/s\]

The time for the parachutist to reach the ground is from equation 3.

\[300m = (2.0m/s)t = 150s\]

When the canister strikes the ground, the parachutist has dropped:

\[(2.0m/s)7.6 = 15m\]

And is 285 m above the ground. The expression for the distance between the canister and the parachutist is:

\[x_{c} - x_{p} = (2.0m/s)t + (4.9m/s^{2})t^{2} - (2.0m/s)t = (4.9m/s^{2})t^{2}\]

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**Example 7**

A coconut is dropped from a height of 60m. One second later, a second coconut is thrown down with an initial velocity. Both coconuts reach the ground at the same time. What was the initial velocity of the second coconut?

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In problems where there is a time delay, it is usually best to calculate the position, velocity, and acceleration of the first particle at the time when the second particle starts to move. It is possible to do time delay problems with a time differential in one set of equations. The difficulty with this approach is that it is easy to get an algebraic sign wrong. If you say that the time for the second particle is the time for the first particle plus the difference between them, then it is essential that the algebraic sign of the difference be correct. It is much easier to do them, in this slower but inherently more accurate way. First, calculate the state of the first particle when the second one begins moving. Then write the two sets of equations describing the motion with this instant as t=0.

Calculate the position and velocity of the first coconut at the end of 1s, the time when the second one starts.

\[x = (½)at^{2} = (½)(9.8m/s^{2})(1.0s)^{2} = 4.9m\]

\[v = at = 9.8m/s^{2}(1.0s) = 9.8m/s\]

Since position, velocity, and acceleration is positive down, orient the coordinate system for positive down with the origin at the top. At the instant the second coconut is thrown down, the first coconut has position 4.9m, velocity 9.8m/s and acceleration 9.8m/s^{2}.

Since both coconuts strike the ground at the same time, use the conditions of the first coconut to find the total time. First, calculate the velocity at impact of the first coconut.

\[v_{1}^{2} = v_{1o}^{2} + 2a(x_{1}-x_{1o}) = (9.8m/s)^{2} + 2(9.8m/s^{2})(60-4.9)m = 1176m^{2}/s^{2} = 34.3m/s\]

The time for the second coconut to reach the ground is the same as the time for the first coconut to go from 4.9 m to 60 m or the time for the first coconut to go from 9.8 m/s to 34.3 m/s.

This comes from \(v = v_{0} + at\) where the velocity of the first coconut when the second one is thrown down, and v is the velocity of the first coconut at the ground.

\[34.3m/s = 9.8 m/s _ (9.8m/s^{2})t = 2.5s\]

Now that we have the time for the second coconut to travel the 60 m, we can find its initial velocity from \(x-x_{0}=v_{2o}t + (½)at^{2}\) where t is the total time for the second coconut.

\[60m = v_{2o}(2.5s) + (4.9m/s^{2}(2.5s)^{2} =11.75m/s\]

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**Example 8**

A boat is passing under a bridge. The deck of the boat is 15 m below the bridge. A small package is to be dropped from the bridge onto the deck of the boat when the boat is 25 m from just below the drop point. What boat speed is necessary to have the package land in the boat?

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Calculate the time for the package to fall the 15.0 m using \(x-x_{0} = v_{o}t+(½)at^{2}\)

\[15m = (4.9m/s^{2})t^{2} = 1.7s\]

The boat must move forward at 25 m/ 1.7s = 14 m/s.

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**Example 9**

You are observing steel balls falling at a constant velocity in a liquid filled tank. The window you are using is 1m high, and the bottom of the window is 12m from the bottom of the tank. You observe a ball falling past the window taking 3.0s to pass the window. Calculate the time required to reach the bottom of the tank after the ball has reached the bottom of the window.

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The observed velocity is \(1.0m/3.0s so 12 m = (1.0m/3.0st = 36s\)

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**Example 10**

In a situation similar to the last example, the tank is filled with a different liquid, causing the acceleration in the tank to be \(6.0 m/s^{2}\) and the time to traverse the window .40 s. Calculate the height of the liquid above the window, the time to reach the bottom of the tank, and the velocity of the ball when it reaches the bottom of the tank.

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From the data about the window, calculate the velocity of the ball at the top of the window.

\[x-x_{o} = v_{t}t + (½)at^{2}\]

\[1.0m = v_{t}(.40s) + (3.0m/s^{2})(.40s)^{2} = 1.3m/s\]

The velocity at the bottom of the window is from:

\[v_{b}^{2} = v_{t}^{2} + 2a(x-x_{o}) = (1.3m/s)^{2} + 2(6.0m/s^{2})1.0m = 3.7m/s\]

The time to reach the bottom of the tank is from \(x-x_{o} = v_{b}t + (½)at^{2}\)

\[12m = (3.7m/s)t + (3.0m/s^{2})t^{2}\]

Eliminate the units

\[t = \frac{-3.7\pm \sqrt{(3.7)^{2} - 4(3.0)(-12)}}{2(3)} = \frac{-3.7\pm12.6}{6} = 1.5\]

This positive time is for the ball to travel from the bottom of the window to the bottom of the tank.

Assuming that the ball started at zero velocity, the distance from the top of the liquid to the top of the window will come from \(v_{t}^{2}=v_{o}^{2}+2ax_{t}\)

\[(1.3m/s)^{2} = 2(6.0m/s)x_{t} = .14m\]

The velocity with which the ball strikes the bottom of the tank is:

\[v = v_{b} + at = (3.7m/s) + (6.0m/s^{2})1.5s = 12.7m/s\]

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**Example 11**

A ball is observed to pass a 1.4m tall window going up and later going down. The total observation time is .40s(.20s going up and .20s going down). How high does the ball rise above the window?

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This is another example of a question that seems totally unrelated to the information in the problem. When you don’t know where to start and you do not see a clear path to the desired answer, simply start where you can, calculating what you can and hopefully learning enough to answer the specific question.

One of the first things we can calculate in this problem is the average velocity at the middle of the window.

\[v_{avg} = (1.4m/2.0s) = 7.0m/s\]

This average velocity is the velocity of the ball on the way up and on the way down at the middle of the window. Add this feature to the problem. With this information, we can use \(v^{2} = v_{o}^{2} + 2ax_{t}\) to find the distance the ball rises above the midpoint of the window.

\[(7.0m/s)^{2} = 2(9.8m/s^{2})x_{t} = 2.5m\]

So the ball rises 2.5-0.7=18 m above the top of the window.

View the ball as decelerating as it goes up past the window, ands find \(v_{b}\) at the bottom of the window from \(x-x_{o}=v_{o}t+(½)at^{2}\).

\[1.4m = v_{b}(.20s)-(4.9m/s^{2})(.20s)^{2} = 8.0m/s\]

Velocity and displacement are taken as positive, so acceleration is negative.

Now the distance to maximum height, where velocity is zero, is:

\[(8.0m/s)^{2} = 2(9.8m/s^{2})x_{m} = 3.2m\]

Again, the maximum height of the ball above the top of the window is 3.2 m - 1.4 m = 1.8m.