Question

A ball falls on an inclined plane of inclination $\theta$ from a height h above the point of impact and makes a perfectly elastic collision.Where will it hit the plane again ?

Answer 1

The ball strikes the inclined plane at origin with velocity $v_0=\sqrt{2gh}$.As the ball elastically rebounds,it recalls with same velocity $v_0$,at the same angle $\theta$ from the normal or y-axis.Let the ball strikes the incline second time at any point P,which is at a distance l from the origin along the incline.From the equation,

$y=v_{0y}t+\frac{1}{2}{w}_y{t}^2$

$0=v_{0}cos\theta t-\frac{1}{2}gcos\theta t^2$

Which t is the same time of motion of ball in air which moving from origin to P.

As $t\ne 0,sot=\frac{2v_0}{g}$

Now from the equation $x=v_{0x}t+\frac{1}{2}{w}_x{t}^2$

$l=v_{0}sin\theta t+\frac{1}{2}gsin\theta t^2$

So, $l=v_{0}sin\theta \left(\frac{2v_0}{g}\right)+\frac{1}{2}gsin\theta {\left(\frac{2v_0}{g}\right)}^2$

$=\frac{4v_0^{2}sin\theta }{g}$

Hence the plane will hit again at a distance $l=8hsin\theta$