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## Homework Statement

A tennis ball with a velocity of +9.6 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 -5.2 m/s (to the left). If the ball is in contact with the wall for 0.008 s, what is the average acceleration of the ball while it is in contact with the wall?

Vi=9.6m/s

Vf=-5.2m/s

t=.008s

## Homework Equations

Aavg = Vf-Vi / Tf-Ti

Vf = Vi+a(Tf-Ti)

## The Attempt at a Solution

9.6-5.2 / .008 = 550 m/s^2 left ???

I realize that answer is nowhere close to correct. I must be using an incorrect formula because its dealing with the wall. If anyone could guide me to a correct formula that would be awesome!