A ball is dropped from a height h on a floor if the coefficient of restitution ise. Find the
(a). Speed of ball after the first second … `n^(th)` collision.
(b). Maximum height attained by the ball, after the first, second … `n^(th)` collision.
(c). Time taken by the ball to reach the highest point after the first, second, ...`n^(th)` collision.
(d). total distance covered by the ball.
(e). totol time of journey.

Explanation

O to A:
Speed of the ball when it hits the ground `u=sqrt(2gh)`
Time take by the ball to hit the floor,
`h=(1)/(2)g t_(0^(2))impliest_(0)=sqrt((2h)/(g))`
`v_(1)-v_(2)=-e(u_(1)-u_(2))`
1 ball, 2 floor
`u_(2)=v_(2)=0`
`v_(1)=-eu_(1)=-eu`
Speed of the ball after the first collision `=eu`
A to B
`0=v_(1)^(2)-2gh_(1)impliesh_(1)=(v_(1)^(2))/(2g)=(e^(2)u^(2))/(2g)=(e^(2)xx2gh)/(2g)=e^(2)h`
Height attained by the ball after the first collision `=e^(2)h`
`0=v_(1)-g t_(1)impliest_(1)=(v_(1))/(g)=(eu)/(g)=(e)/(g)sqrt(2gh)=esqrt((2h)/(g))=et_(0)`
After every collision the speed becomes e times. Maximum height becomes `e^(2)` times. time to reach at the heighest point becomes e times.
`{:(,1^(st)"collision","After 2nd collision",n^(th)" collsion"),("speed of ball",eu,e^(2)u,e^(n)u),("Maximum height",e^(2)h,e^(4)h,e^(2n)h),("Time to reach at highest point",et_(0),et_(0),e^(n)t_(0)):}`
when `u=sqrt(2gh),t_(0)=sqrt(2h//g)`
Total distance covered by the ball
`S=h+2h_(1)+2h_(2)+. . . infty`
`=h+2e^(2)h+2e^(4)h+. . . infty`
`=h+2e^(2)h{1+e^(2)+. . .infty}`
`=h+2e^(2)h{(1)/(1-e^(2))}=((1+e^(2))/(1-e^(2)))h`
Total time of journey
`T=t_(0)+2t_(1)+2t_(2)+... infty`
`=t_(0)+2et_(0)+2e^(2)t_(0)+. . . infty`
`=t_(0)+2et_(0)(1+e+ . . .infty)`
`=t_(0)+2et_(0)((1)/(1--e))=((1+e)/(1-e))t_(0)=((1+e)/(1-e))sqrt((2h)/(g))`