A tennis ball with a speed of 15.7 m/s is moving perpendicular to a wall. After striking the wall, the ball rebounds in the opposite direction with a speed of 14.444 m/s. If the ball is in contact with the wall for 0.0099 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 .

-3044.848m/s^2

Explanation:

Initial velocity = 15.7 m/s (u)

Final velocity = -14.444 m/s (v) velocity after striking moving opposite direction.

Time = 0.0099 s

Average acceleration :

change in velocity/ total time

= V - U/t

= -14.444 m/s - 15.7 m/s/0.0099 s

= -30.144/0.0099 s

= -3044.848m/s^2

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