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

Relevance
  • 1 decade ago
    Favorite 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)

Still have questions? Get your answers by asking now.