Kinematic Equations for Objects in Free Fall

Example

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.

Strategy

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 $$ {y}_{0}=0,{v}_{0}=-4.9\,\text{m/s,}\,\text{and}\,g=9.8\,{\text{m/s}}^{2}$$.

Solution

1. Show Answer

   Substitute the given values into the equation: $$ \begin{array}{cc} y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}\hfill \\ -98.0\,\text{m}=0-(4.9\,\text{m/s})t-\frac{1}{2}(9.8\,{\text{m/s}}^{2}){t}^{2}.\hfill \end{array} $$ This simplifies to $$ {t}^{2}+t-20=0. $$ This is a quadratic equation with roots $$ t=-5.0\mathrm{s}\,\text{and}\,t=4.0\mathrm{s}$$. The positive root is the one we are interested in, since time $$ t=0 $$ is the time when the ball is released at the top of the building. (The time $$ t=-5.0\mathrm{s} $$ 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.)
2. Show Answer

   Using (Figure), we have $$ v={v}_{0}-gt=-4.9\,\text{m/s}-(9.8{\text{m/s}}^{2})(4.0\,\text{s})=-44.1\,\text{m/s}\text{.}$$

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 $$ t=0 $$ 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.

Example

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.

Strategy

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.

Solution

1. (Figure) gives
   $$y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}$$
   $$0=0+{v}_{0}(5.0\,\text{s})-\frac{1}{2}(9.8\,\text{m}\text{/}{\text{s}}^{2}){(5.0\,\text{s})}^{2},$$

   which gives $$ {v}_{0}=24.5\,\text{m/sec}$$.

2. At the maximum height, $$ v=0$$. With $$ {v}_{0}=24.5\,\text{m/s}$$, (Figure) gives
   $${v}^{2}={v}_{0}^{2}-2g(y-{y}_{0})$$
   $$0={(24.5\,\text{m/s})}^{2}-2(9.8{\text{m/s}}^{2})(y-0)$$

   or

   $$y=30.6\,\text{m}\text{.}$$
3. To find the time when $$ v=0$$, we use (Figure):
   $$v={v}_{0}-gt$$
   $$0=24.5\,\text{m/s}-(9.8{\text{m/s}}^{2})t.$$

   This gives $$ t=2.5\,\text{s}$$. Since the ball rises for 2.5 s, the time to fall is 2.5 s.

4. 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.
5. Show Answer

   The velocity at $$ t=5.0\mathrm{s} $$ can be determined with (Figure): $$ \begin{array}{cc}\hfill v& ={v}_{0}-gt\hfill \\ & =24.5\,\text{m/s}-9.8{\text{m/s}}^{2}(5.0\,\text{s})\hfill \\ & =-24.5\,\text{m/s}.\hfill \end{array}$$

Significance

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.

Example

Rocket Booster

A small rocket with a booster blasts off and heads straight upward. When at a height of $$ 5.0\,\text{km} $$ 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?

Strategy

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.

Solution

1. From (Figure),
   Show Answer

   $$ {v}^{2}={v}_{0}^{2}-2g(y-{y}_{0})$$. With $$ v=0\,\text{and}\,{y}_{0}=0$$, we can solve for y: $$ y=\frac{{v}_{0}^{2}}{-2g}=\frac{(2.0\,×\,{10}^{2}\text{m}\text{/}{\text{s})}^{2}}{-2(9.8\,\text{m}\text{/}{\text{s}}^{2})}=2040.8\,\text{m}\text{.} $$ 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.
2. Show Answer

   An altitude of 6.0 km corresponds to$$y=1.0\,×\,{10}^{3}\,\text{m} $$ in the coordinate system we are using. The other initial conditions are$${y}_{0}=0,\,\text{and}\,{v}_{0}=200.0\,\text{m/s}$$.

   We have, from (Figure),

   Show Answer

   $${v}^{2}={(200.0\,\text{m}\text{/}\text{s})}^{2}-2(9.8\,\text{m}\text{/}{\text{s}}^{2})(1.0\,×\,{10}^{3}\,\text{m})⇒v=±142.8\,\text{m}\text{/}\text{s}.$$

Significance

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.