A ball starts falling freely from a height h from a point on the inclined plane forming an angle `alpha` with the horizontal as shown. After collision with the incline it rebounds elastically off the plane. Then it again strikes the incline at

Explanation

Let ball strike the plane with speed u .
` u = sqrt(2gh) " "…(i)`
Components of this velocity along the plane and perpendicular to the plane are shown in figure .

After first collision with the plane , component of velocity `( u sin theta )` perpendicular to the plane becomes `eu cos theta ` after the impact and also it reverse its direction . Let X - axis is along the palne and Y - axis is perpendicular to the palne as shown in figure . Components of acceleration due to gravity are also shown in the figure . Let the ball strike the plane at a distance R from the point of impact . So displacement along X - axis becomes zero when ball again strikes the plane .
`S_(y) = u_(y)t+1/2 a_(y)t^(2)rArr 0 = ( eu cos theta ) t + 1/2 (- g cos theta ) t^(2)`
` rArr " " t - (2cu)/g " "...(ii)`
`S_(x) =u_(x)t +1/2 a_(x)t^(2) rArr R = ( u sin theta ) t +1/2 ( g sin theta ) t^(2)`
Substitting from equation (ii) we get
` rArr " " R = ( u sin theta ) ((2eu)/g) +1/2 (g sin theta ) ((2eu)/g)^(2)`
`rArr " " R = (2eu^(2)sin theta)/g + (2e^(2)u^(2)sin theta )/g `
`rArr " "R = (2e(1+e)u^(2)sin theta )/g `
Substituting from equation (i) we get
` rArr " "R = (2e(1+e)(2gh )sin theta )/g `
`rArr R = 4e (1+e) h sin theta `