Free Fall Motion

An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is said to be in free fall. If the height through which the object falls is small compared to the earth’s radius, g can be taken to be constant, equal to 9.8 m$s^{–2}$. Free fall is thus a case of motion with uniform acceleration.

We assume that the motion is in y-direction, more correctly in –y-direction because we choose upward direction as positive. Since the acceleration due to gravity is always downward, it is in the negative direction and we have

$a=-g$ = 9.80 $m.s^{-2}$

The object is released from rest at y = 0. Therefore, $v_0$ = 0 and the equations of motion become:

v = 0 – gt = –9.8t m$s^{–1}$ 

y = 0 – ½ g$t^2$ = –4.9$t^2$ m 

$v^2$ = 0 – 2gy = –19.6y $m^2.s^{–2}$ 

Fig. 1: Motion of an object under free fall. (a) Variation of acceleration with time. (b) Variation of velocity with time. (c) Variation of distance with time

These equations give the velocity and the distance travelled as a function of time and also the variation of velocity with distance. The variation of acceleration, velocity, and distance, with time have been plotted in Fig. 1(a), (b) and (c).

Example 1

Galileo’s law of odd numbers : “The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7…...].” Prove it.

Answer 

Let us divide the time interval of motion of an object under free fall into many equal intervals τ and find out the distances traversed during successive intervals of time. Since initial velocity is zero, we have

y = $-\frac{1}{2}gt^2$

Using this equation, we can calculate the position of the object after different time intervals, 0, τ, 2τ, 3τ… which are given in second column of Table 1. If we take (–1/2) g$τ^2$ as $y_0$ — the position coordinate after first time interval τ, then third column gives the positions in the unit of $y_o$. 

Table 1

The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11… as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall.

Example 2

Reaction time: When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on complexity of the situation and on an individual. 

Fig.2: Measuring the reaction time.

You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger (Fig. 2). After you catch it, find the distance d travelled by the ruler. In a particular case, d was found to be 21.0 cm. Estimate reaction time.

Answer 

The ruler drops under free fall. Therefore, $v_o$  = 0, and a = – g = –9.8 m$s^{–2}$. The distance travelled d and the reaction time tr are related by

$d=-\frac{1}{2}gt_r^2$

Or, $t_r=\sqrt{\frac{2d}{g}}$

Given d = 21.0 cm and g = 9.8 m$s^{–2}$ the reaction time is

$t_r=\sqrt{\frac{2 \times 0.21 \ m}{9.80 \ m.s^{-2}}}$ = 0.2 s

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