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.

Answer 1

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

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, 

Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as: v = u_r + at′

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 

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: