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

A ball starts falling with zero initial velocity on a smooth inclined

1.	A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle alpha with the horizontal. Having fallen the distnace 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» Solution :The SITUATION is shown in fig Let the ball rebound from point A and return to the plane at point B, MOVING along the trajectory shown in diagram. Let `AB=l` Velocity of rebound at point `A,V=sqrt(2gh)` Velocity COMPONENT perpendicular to plane `=v cos alpha` velocity component along the plane `=v sin alpha` Retardation perpendicular to plane `=g cos alpha` Acceleration along the plane `=g sin alpha` If the ball comes to REST at this highest point after time, t then `(v cos alpha-g cos alpha.t)=0` or `t=v/g` Time of fight of the ball `T=2t=(2v)/g` Range of the ball along the plane `l=(vsin alpha)T+1/2(g sin alpha)T^(2)` `=v sin alpha.(2v)/g+1/2. g sin alpha(4v^(2))/(g^(2))` `=(2v^(2))/g sin alpha+(2v^(2))/g sin alpha` `=4(v^(2))/g sin alpha=4((2gh))/gsin alpha` `( :. v=sqrt(2gh))"":. l=8h sin alpha`