Question

A tennis ball with a speed of 11.3 m/s is moving perpendicular to a wall. after striking the wall, the ball rebounds in the opposite direction with a speed of 7.2207 m/s. if the ball is in contact with the wall for 0.00803 s, what is the average acceleration of the ball while it is in contact with the wall? take "toward the wall" to be the positive direction. answer in units of m/s 2 .

Answer 1

To calculate the average acceleration of the ball we use the formula,

$a= \frac{v_{f} -v_{i}}{t}$

Here, $v_{f}$ is the final velocity of the ball and $v_{i}$ is the initial velocity of the ball t is the time in contact with the wall.

Given $v_{i} = 11.3 m/s$ towards the wall and $v_{f} = 7.2207 m/s$

away from the wall and $t=0.00803 s$.

Substituting these values in above formula , we get

$a=\frac{(-7.2207 m/s)-11.3 m/s}{0.00803 s}$

Here$v_{f}$ is negative because ball is moving away from the wall.

$a= - 2306 .4 \ m/s^{2}$

Therefore, average acceleration of the ball is $- 2306 .4 \ m/s^{2}$ (away from the wall).