A ball is dropped from a height of a height of 90 m on a floor. At each collsion 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

As ball is dropped downwards.
Therefore, `u=0,a=10 ms^(-2),s=90m`
v?, t=?
`s=ut+(1)/(2)at^(2)`
`=0+(1)/(2)at^(2)` [Here a=g= `10m//s^(2)`]
`rArr t-sqrt((2s)/(a))=sqrt((2xx90)/(10))=4.24` s
`v^(2)-u^(2)=2as`
`v^(2)-0=2as`
`v=sqrt(2 as)`
`=sqrt(2xx10xx90)=30sqrt(2)`
Rebound velocity `=v=(9)/(10)v`
`=(9)/(10)xx30sqrt(2)=27sqrt(2) m//s`
t=time taken to reach highest point `=(v)/(10)=(27sqrt(2))/(10)`
`=2.7sqrt(2)=3.81s`
Total time taken by ball `=t+t=4.24+3.81=8.05`s
As time of ascent=time of descent.
`therefore` The ball will take 3.81 s to come back to floor
Velocity of the ball before striking the floor
`=27sqrt(2) m//s`
Velocity of the ball after striking the floor
`=27sqrt(2)xx(9)/(10)`
`=24.3sqrt(2)` m/s
`therefore` Total time elapsed for `2^("nd")` rebound `=8.05+3.81`
=11.86s