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?

Explanation

On impact at point P. the velocity of the ball,`v = sqrt(2gh)`. Since the ball rebounds elastically, it gets deflected with the same velocity at an angle 8. From the geometry. `theta=a`The distance (PP) where it makes second impact is its range. The motion of the particle along the plane has speed v sin a and acceleration ng sind. The motion along the direction normal to the plane is with velocity v cos a and acceleration - cosa. Hence the time of flight T is given by y=0
. y = 0=v cos`aT-(1)/(2)gcosaT^(2)`or`T=(2v)/(g)`
Now l=vsina`T+(1)/(2)gsinaT^(2)`