A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle `alpha` with the horizontal. Having fallen the distance h, the ball rebounds elastically off the inclined plane. At what distance from the impact point will the ball rebound for the second time?

Answer

`l=8h sin alpha`

Explanation

The ball strikes the inclined plane `(Ox)` at point O (origin) with velocity `v_0=sqrt(2gh)` (1)
As the ball elastically rebounds, it recalls with same velocity `v_0`, at the same angle `alpha` from the normal or y axis (figure). Let the ball strikes the incline second time at P, which is at a distance `l`(say) from the point O, along the incline. From the equation
`y=v_(0y)t+1/2w_yt^2`
`0=v_0cos alpha tau-1/2gcos alpha t^2`
where `tau` is the time of motion of ball in air while moving from O to P.
As `tau!=0`, so, `tau=(2v_0)/(g)` (2)
Now from the equation.
`x=v_0xt+1/2w_xt^2`
`l=v_0sin alpha tau+1/2 g sin alpha tau^2`
so, `l=v_0 sin alpha((2v_0)/(g))+1/2 g sin alpha ((2v_0)/(g))`
`=(4v_0^2sin alpha)/(g)` (using 2)
Hence the sought distance, `l=(4(2gh)sin alpha)/(g)=8h sin alpha` (Using Eq. 1)