Question

A tennis ball with a velocity of +10.0 m/s to the right is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a velocity of â8.0 m/s to the left. If the ball is in contact with the wall for 0.012 s, what is the average acceleration of the ball while it is in contact with the wall?

Answer 1

Answer:

-1500 m/s2

Explanation:

So the ball velocity changes from 10m/s into the wall to -8m/s in a totally opposite direction within a time span of 0.012s. Then we can calculate the average acceleration of the ball as the change in velocity over a unit of time.

$a = \frac{\Delta v}{\Delta t} = \frac{-8 - 10}{0.012} = \frac{-18}{0.012} = -1500 m/s^2$