Question

# A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.

Solution

## Ball is dropped from a height, s = 90 m Initial velocity of the ball, u = 0 Acceleration, a=g=9.8 m/s2 Final velocity of the ball = v From second equation of motion, time (t) taken by the ball to hit the ground can be obtained as: s=ut+12at290=0+12×9.8t2t=√18.38=4.29 s From first equation of motion, final velocity is given as: v = u + at =0+9.8×4.29=42.04 m/s Rebound velocity of the ball, ur=910v=910×42.04=37.84 m/s Time (t) taken by the ball to reach maximum height is obtained with the help of first quation of motion as: v = u' + at' 0=37.84+(−9.8)t′⇒t′=−37.84−9.8=3.86 s Total time taken by the ball = t + t' = 4.29 + 3.86 = 8.15 s As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time. The velocity with which the ball rebounds from the floor =910×37.84=34.05 m/s Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s The speed-time graph of the ball is represented in the given figure as:

Suggest corrections