A ball is dropped from a height of 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.

Explanation

Ball is dropped from a height, s = 90 m
Initial velocity of the ball, u = 0
Acceleration, a = g = `9.8 m//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+1/2at^2`
`90=0+1/2xx9.8t^2`
`t=sqrt(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, `u_r =9/10v=9/10xx42.04=37.84 m//s`
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′`
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 `=9/10xx37.84`=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: