A ball is dropped from a height of 1m onto a horizontal plane. The ball takes 1.3 s from its time of release for the second impact with the plane. Find the coefficient of restitution.

Explanation

Let the velocity of the ball be V just before the 1st impact with the plane.
Then` V^2=2gh=2xx980xx100`
`or, V=sqrt(2xx980xx100)=442.7 cm*s^(-1)`
If the time taken by the ball is t, to fall from the height of 1m ,then
`t=sqrt((2h)/(g))` [using `h=1/2 "gt"^2`]
`=sqrt(2xx(100)/(980))=0.452 s.`
so, time interval between the first and the second impacts =1.3-0.452=0.848 s.
Time taken by the ball to reach the highest point after the first impact is,
`t_1=1/2xx0.848=0.424 s`
If the velocity of separation of the ball after the first impact is v, then
0= v-`"gt"_1`
`or, v="gt"_1 =980xx0.424=415.5 cm*s^(-1)`
`therefore` Coefficient of restitution,
`e=v/V=(415.5)/(442.7)=0.94`