3.5 Free Fall

Gravity

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 3.26.

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/s² to 9.83 m/s², depending on latitude, altitude, underlying geological formations, and local topography, let’s use an average value of 9.8 m/s² 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 $a=\text{−}g=\mathrm{-9.8}{\text{m/s}}^2,$ and if we define the downward direction as positive, then $a=g=9.8{\text{m/s}}^2$.

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.

Example 3.14

Free Fall of a Ball

Figure 3.27 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 Equation 3.16, with $y_0=0,v_0=\mathrm{-4.9}\text{m/s,}\text{and}g=9.8{\text{m/s}}^2$.

Solution

1. Substitute the given values into the equation:
   $$\begin{array}{}\\ \\ y=y_0+v_{0}t-\frac{1}{2}g{t}^2\hfill \\ -98.0\text{m}=0-(4.9\text{m/s})t-\frac{1}{2}(9.8{\text{m/s}}^2)t^2.\hfill \end{array}$$
   This simplifies to
   $$t^2+t-20=0.$$
   This is a quadratic equation with roots $t=\mathrm{-5.0}\mathrm{s}\text{and}t=4.0\mathrm{s}$. 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 $t=\mathrm{-5.0}\mathrm{s}$ 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. Using Equation 3.15, we have
   $$v=v_0-gt=\mathrm{-4.9}\text{m/s}-(9.8{\text{m/s}}^2)(4.0\text{s})=\mathrm{-44.1}\text{m/s}\text{.}$$

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 3.15

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 3.28. (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. Equation 3.16 gives
   $$y=y_0+v_{0}t-\frac{1}{2}g{t}^2$$
   $$0=0+v_0(5.0\text{s})-\frac{1}{2}\left(9.8\text{m}\text{/}{\text{s}}^2\right){(5.0\text{s})}^2,$$
   which gives $v_0=24.5\text{m/s}$.
2. At the maximum height, $v=0$. With $v_0=24.5\text{m/s}$, Equation 3.17 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 Equation 3.15:
   $$v=v_0-gt$$
   $$0=24.5\text{m/s}-(9.8{\text{m/s}}^2)t.$$
   This gives $t=2.5\text{s}$. Since the ball rises for 2.5 s, the time to fall is 2.5 s.
4. 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. The velocity at $t=5.0\mathrm{s}$ can be determined with Equation 3.15:
   $$\begin{array}{cc}\hfill v& =v_0-gt\hfill \\ & =24.5\text{m/s}-9.8{\text{m/s}}^2(5.0\text{s})\hfill \\ & =\mathrm{-24.5}\text{m/s}.\hfill \end{array}$$

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 that 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 3.16

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 Equation 3.17, which gives us the maximum height of the booster. We also use Equation 3.17 to give the velocity at 6.0 km. The initial velocity of the booster is 200.0 m/s.

Solution

1. From Equation 3.17, $v^2=v_0^2-2g(y-y_0)$. With $v=0\text{and}{y}_0=0$, we can solve for y:
   $$y=\frac{v_0^2}{2g}=\frac{(2.0\times {10}^2\text{m}\text{/}{\text{s})}^2}{2(9.8\text{m}\text{/}{\text{s}}^2)}=2040.8\text{m}\text{.}$$
   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. An altitude of 6.0 km corresponds to$y=1.0\times {10}^3\text{m}$ 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 Equation 3.17,
   $$v^2={(200.0\text{m}\text{/}\text{s})}^2-2(9.8\text{m}\text{/}{\text{s}}^2)(1.0\times {10}^3\text{m})\Rightarrow v=\pm 142.8\text{m}\text{/}\text{s}.$$

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.