Question

4.A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle α (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 1

see the diagram.

The ball falls from A on the to incline and bounces off at B.  BB' is the normal to the incline at B.  Since it is a smooth plane and collision/bouncing is elastic, the ball bounces at the same angle with the normal BB'.  The velocity remains same.

Hence the angle of projection of the ball wrt horizontal = 90° - 2α.
Velocity of the ball at B = v = √(2gh).

Let B= (0,0).

The equation of the trajectory of the ball (parabolic path) :
$ball:\ \ y = x*tan (90^0-2\alpha) - \frac{g x^2}{2v^2} Sec^2(90^0-2 \alpha)\\\\Equation\ of\ inclined\ plane:\ y=-x*tan\ \alpha\\\\Ball\ hits\ the\ plane\ at:\ \ -x*tan\alpha=x*cot(2\alpha) -\frac{g x^2}{4gh} cosec^2(2 \alpha)\\\\so\ x=0 \ \ or, \ \ x=(cot\ 2\alpha+tan\ \alpha)4h \ Sin^2(2\alpha)\\\\or,\ x=\frac{4h*(1-tan^2\alpha+2tan^2\alpha)Sin^2 2\alpha}{2\ tan\alpha}\\\\x=4h\ Sin\ 2\alpha\\\\y=-4h\ sin\ 2\alpha*tan\ \alpha=-8h\ sin^2\alpha\\\\Distance=\sqrt{x^2+y^2}=8h\ Sin\alpha,}$

OR, the Distance on the inclined plane from B (point of 1st bounce to point of 2nd bounce) =  x / Cos α = 8 h Sin α..

Answer 2

Explanation:

according to the question answer is given