Question

A tennis ball is dropped on the floor from a height of 20 m. It rebounds to a height of 5 m. The ball was in contact with the floor for 0.01 s. What was its average acceleration during the contact? (g = 10 m/s$^2$)

A. -3000 m/s$^2$

B. -2000 m/s$^2$

C. -1000 m/s$^2$

D. 500 m/s$^2$

Answer 1

Hint: By definition, acceleration for a body if the difference of its final and initial velocity divided by the time in which the velocity changes occurs. It is the rate of change of velocity. 

Formula used:

Third law of motion:

$v^2 - u^2 = 2 a s$.

Acceleration by definition is written as,

$a = \dfrac{v - u}{t}$.

Complete answer:

The problem has to be divided into two parts, one when the free fall from the height of 20 m occurs and two, when the ball rebounds to a height of 5 m. Since in both the cases, one of the velocities (initial or final) is zero, we will get our required velocity before 0.01 s and after 0.01 s, so that we can find the acceleration.

1. The ball falls freely from height s = 20 m height with u = 0, so we may write 

$v^2 - 0 = 2 \times 10 \times 20$

$\implies v = 20$ m/s.

Now, at the end of 20 m distance, i.e., just as the ball reaches the ground, it has a velocity 20 m/s.

2. As ball rises to a height s = 5 m, v = 0, so we may write:

$ - u^2 = - 2 \times 10 \times 5$

$\implies u = 10$ m/s.

As direction of velocity is upward during this step so we can take $u = -10\; m/s$

This will be the velocity of the ball from the ground, so that it will be able to reach a height of 5 m.

Now, the u that we found out in case 2 is the final velocity of the ball and the v that we found in case 1 is actually the initial velocity of the ball, so by using acceleration formula, we get:

$a = \dfrac{-10 - 20}{0.01} = - 3000 m/s^2$.

So, the correct answer is “Option C”.

Note:

Be careful with the positive and negative signs everywhere. Since g is in the downward direction, we always write it with a negative sign. The distance travelled from some height to the ground are taken negative and from ground to some height are taken positive. Also the acceleration to be found will have a negative sign as gravity is retarding the motion of the ball.