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# Help with Physics question!!!?

A tennis ball with a velocity of 50.0m/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 -8m/s to the left. If the ball is in contact with the wall for 0.014s, what is the average acceleration of the ball while it is in contact with the wall?

Answer should be in m/s^2

### 1 Answer

- 1 decade agoFavorite Answer
Shaik Here:

Velocity of Tennis Ball,Vb=50.0m/s

Rebounce Velocity,Vrb=-8m/s

Time on Wall,t=0.014s

Average Acceleration, Aavg=?

As we know that

Acceleration, a = distance S by Time, t.

a=s/t or a=m/s^2.(Acceleration = Meter/Second^2)

What to do?

Acceleration While Hitting the Wall, avb = Vb/t

avb=50/0.014^2

avb=255102.04 m/s^2

Now Rebound Acceleration, avrb = Vrb/t

avrb = -8/0.014^2

avrb = -40816.32

Going towrds the Average Acceleration, Aavg = ????

Aavg = avb+avrb/2

Aavg = 255102.04+ (-40816.32)/2

Aavg = 214285.72/2

Aavg = 107142.86 m/s^2 (Required Answer)

Thank You

(sha_ik15@yahoo.com)