2.5 Free Fall

The motion of an object moving under the influence of gravity only, and no other forces, is called free fall. Strictly speaking, free fall occurs only in a vacuum, where there is no air resistance. Fortunately, the effect of air resistance is small for “heavy objects,” so we’ll make only a very slight error in treating these objects as if they were in free fall. For very light objects, such as a feather, or for objects that fall through very large distances and gain very high speeds, the effect of air resistance is not negligible. Motion with air resistance is a problem we will study in Chapter 6. Until then, we will restrict our attention to “heavy objects” and will make the reasonable assumption that falling objects are in free fall.

Galileo, in the 17th century, was the first to make detailed measurements of falling objects. The story of Galileo dropping different weights from the leaning bell tower at the cathedral in Pisa is well known, although historians cannot confirm its truth. Based on his measurements, wherever they took place, Galileo developed a model for motion in the absence of air resistance:

Two objects dropped from the same height will, if air resistance can be neglected, hit the ground at the same time and with the same speed.
Consequently, any two objects in free fall, regardless of their mass, have the same acceleration $\vec{a}_{free fall}$ .

A time-lapse photograph taken at 6 instances of time shows that an apple and a feather fall at the same rate in vacuum. The distance between the instances of the falling apple as well as that of the falling feather increases gradually from top to bottom. — In a vacuum, the apple and feather fall at the same rate and hit the ground at the same time.

FIGURE 2.25a shows the motion diagram of an object that was released from rest and falls freely. FIGURE 2.25b shows the object’s velocity graph. The motion diagram and graph are identical for a falling pebble and a falling boulder. The fact that the velocity graph is a straight line tells us the motion is one of constant acceleration, and $a_{free fall}$ is found from the slope of the graph. Careful measurements show that the value of $\vec{a}_{free fall}$ varies ever so slightly at different places on the earth, due to the slightly nonspherical shape of the earth and to the fact that the earth is rotating. A global average, at sea level, is

The figure labeled ("a") illustrates the motion diagram of a particle in free fall.

The figure labeled (b) illustrates the velocity-versus-time graph for the motion of an object in free fall.

Full Alternative Text

\vec{a}_{free fall} = (9.80 m / s^{2}, vertically downward)

(2.24)

Vertically downward means along a line toward the center of the earth.

The length, or magnitude, of $\vec{a}_{free fall}$ is known as the free-fall acceleration, and it has the special symbol $g$ :

g = 9.80 m / s^{2} (free-fall acceleration)

Several points about free fall are worthy of note:

g, by definition, is always positive. There will never be a problem that will use a negative value for g. But, you say, objects fall when you release them rather than rise, so how can $g$ be positive?
g is not the acceleration $a_{free fall}$ , but simply its magnitude. Because we’ve chosen the y-axis to point vertically upward, the downward acceleration vector $\vec{a}_{free fall}$ has the vertical component

$a_{y} = a_{free fall} = - g$ (2.25)

It is $a_{y}$ that is negative, not $g$ .
We can model free fall as motion with constant acceleration, with $a_{y} = - g$ .
g is not called “gravity.” Gravity is a force, not an acceleration. The symbol $g$ recognizes the influence of gravity, but $g$ is the free-fall acceleration. You may also see $g$ called the acceleration due to gravity.
$g = 9.80 m / s^{2}$ only on earth. Other planets have different values of $g$ . You will learn in Chapter 13 how to determine $g$ for other planets.

NOTE Despite the name, free fall is not restricted to objects that are literally falling. Any object moving under the influence of gravity only, and no other forces, is in free fall. This includes objects falling straight down, objects that have been tossed or shot straight up, and projectile motion.

EXAMPLE 2.12 A falling rock

A rock is dropped from the top of a 20-m-tall building. What is its impact velocity?

MODEL A rock is fairly heavy, and air resistance is probably not a serious concern in a fall of only 20 m. It seems reasonable to model the rock’s motion as free fall: constant acceleration with $a_{y} = a_{free fall} = - g$ .

VISUALIZE FIGURE 2.26 shows the pictorial representation. We have placed the origin at the ground, which makes $y_{0} = 20 m$ . Although the rock falls 20 m, it is important to notice that the displacement is $Δ y = y_{1} - y_{0} = - 20 m$ .

A pictorial representation of a falling rock and its motion diagram.

FIGURE 2.26 Full Alternative Text

SOLVE In this problem we know the displacement but not the time, which suggests that we use the third kinematic equation from Problem-Solving Strategy 2.1:

v_{1 y}^{2} = v_{0 y}^{2} + 2 a_{y} Δ y = - 2 g Δ y

We started by writing the general equation, then noted that $v_{0 y} = 0 m / s$ and substituted $a_{y} = - g$ . Solving for $v_{1 y} :$

v_{1 y} = \sqrt{- 2 g Δ y} = \sqrt{- 2 (9.8 {m/s}^{2}) (- 20 m)} = \pm 20 m / s

A common error would be to say, “The rock fell 20 m, so $Δ y = 20 m$ .” That would have you trying to take the square root of a negative number. As noted above, $Δ y$ is a displacement, not a distance, and in this case $Δ y = - 20 m$ .

The $\pm$ sign indicates that there are two mathematical solutions; therefore, we have to use physical reasoning to choose between them. The rock does hit with a speed of 20 m/s, but the question asks for the impact velocity. The velocity vector points down, so the sign of $v_{1 y}$ is negative. Thus the impact velocity is $- 20 m / s$ .

REVIEW Is the answer reasonable? Well, 20 m is about 60 feet, or about the height of a five- or six-story building. Using $1 m / s \approx 2 mph$ , we see that $20 m / s \approx 40 mph$ . That seems quite reasonable for the speed of an object after falling five or six stories. If we had misplaced a decimal point, though, and found $2.0 m / s$ , we would be suspicious that this was much too small after converting it to $\approx 4 mph.$

EXAMPLE 2.13 Finding the height of a leap

The springbok, an antelope found in Africa, gets its name from its remarkable jumping ability. When startled, a springbok will leap straight up into the air—a maneuver called a “pronk.” A springbok goes into a crouch to perform a pronk. It then extends its legs forcefully, accelerating at $35 m / s^{2}$ for 0.70 m as its legs straighten. Legs fully extended, it leaves the ground and rises into the air. How high does it go?

A photo of a springbok leaping high into the air.

MODEL The springbok is changing shape as it leaps, so can we reasonably model it as a particle? We can if we focus on the body of the springbok, treating the expanding legs like external springs. Initially, the body of the springbok is driven upward by its legs. We’ll model this as a particle—the body—undergoing constant acceleration. Once the springbok’s feet leave the ground, we’ll model the motion of the springbok’s body as a particle in free fall.

VISUALIZE FIGURE 2.27 shows the pictorial representation. This is a problem with a beginning point, an end point, and a point in between where the nature of the motion changes. We’ve identified these points with subscripts 0, 1, and 2. The motion from 0 to 1 is a rapid upward acceleration until the springbok’s feet leave the ground at 1. Even though the springbok is moving upward from 1 to 2, this is free-fall motion because the springbok is now moving under the influence of gravity only.

A pictorial representation of a leaping springbok and its motion diagram.

FIGURE 2.27 Full Alternative Text

How do we put “How high?” into symbols? The clue is that the very top point of the trajectory is a turning point, and we’ve seen that the instantaneous velocity at a turning point is $v_{2 y} = 0$ . This was not explicitly stated but is part of our interpretation of the problem.

SOLVE For the first part of the motion, pushing off, we know a displacement but not a time interval. We can use

\begin{array}{l} v_{1 y}^{2} = v_{0 y}^{2} + 2 a_{0 y} Δ y = 2 (35 {m/s}^{2}) (0.70 m) = 49 m^{2} {/s}^{2} \\ v_{1 y} = \sqrt 49 m^{2} {/s}^{2} = 7.0 m/s \end{array}

The springbok leaves the ground with a velocity of $7.0 m / s$ . This is the starting point for the problem of a projectile launched straight up from the ground. One possible solution is to use the velocity equation to find how long it takes to reach maximum height, then the position equation to calculate the maximum height. But that takes two separate calculations. It is easier to make another use of the velocity-displacement equation:

v_{2 y}^{2} = 0 = v_{1 y}^{2} + 2 a_{1 y} Δ y = v_{1 y}^{2} - 2 g (y_{2} - y_{1})

where now the acceleration is $a_{1 y} = - g$ . Using $y_{1} = 0$ , we can solve for $y_{2}$ , the height of the leap:

y_{2} = \frac{v_{1 y}^{2}}{2 g} = \frac{{(7.0 m/s)}^{2}}{2 (9.80 {m/s}^{2})} = 2.5 m

REVIEW 2.5 m is a bit over 8 feet, a remarkable vertical jump. But these animals are known for their jumping ability, so the answer seems reasonable. Note that it is especially important in a multipart problem like this to use numerical subscripts to distinguish different points in the motion.

2.5 Free Fall

FIGURE 2.25 Motion of an object in free fall.

EXAMPLE 2.12 A falling rock

FIGURE 2.26 Pictorial representation of a falling rock.

EXAMPLE 2.13 Finding the height of a leap

FIGURE 2.27 Pictorial representation of a startled springbok.