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, $a=g=9.8m/s^2$
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:
$s=ut+\frac{1}{2}a{t}^{2}90=0+\frac{1}{2}\times 9.8t^{2}t=\sqrt{18.38}=4.29s$
From first equation of motion, final velocity is given as: v = u + at
$=0+9.8\times 4.29=42.04m/s$
Rebound velocity of the ball, $u_r=\frac{9}{10}v=\frac{9}{10}\times 42.04=37.84m/s$
Time (t) taken by the ball to reach maximum height is obtained with the help of first quation of motion as: v = u' + at'
$0=37.84+(-9.8)t^{\prime }\Rightarrow t^{\prime }=\frac{-37.84}{-9.8}=3.86s$
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 $=\frac{9}{10}\times 37.84=34.05m/s$
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: