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

Explanation

Taking vertical downward motion of ball from a height 90 m
We have
`u=0, a=10m//s^(2), S=90m, t= ?, v=?`
`t=sqrt((25)/a)=sqrt((2 times 90)/10)=3sqrt(25)=4.242`
`V=sqrt(2as)=sqrt(2 times 10 times 30)=30sqrt(2)m//s`

Rebound velocity of ball,
`u^(1)=9/10v=9/10 times 30sqrt(2)=27sqrt(2)m//g`
Time to reach the highest point is
`t^(1)=u^(1)/a=(27sqrt(2))/10=2.7sqrt(2)=3.81 S`
Total time `=t+t^(1)=4.24+3.81=8.05 S`
The ball will take further 3.1S to fall back to floor, where its velocity before striking the floor = `2.7sqrt(2)m//s`.
Time to reach the highest point is,
`t^(1)=u^(1)/a=(27sqrt(2))/(10)=2.7sqrt(2)=3.81S`
Total time `=t+t^(1)=4.24+3.81=8.05S`
The ball will take further 3.81 S fall back to floor, where its velocity before striking the floor `=2.7sqrt(2)m//s`
Velocity of ball after striking the floor
`=9/10times 27sqrt(2)=24.3sqrt(2)m//s`.
Total time elapsed before upward motion of ball.
`=8.05+3.81=11.86 S`
Thus the speed - time graph of this motion will be shown in fig.