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?

Let the ball strikes the inclined plane at origin with velocity $v_0=\sqrt{2gh}$.

form the equation

$y=v_{oy}t+\frac{1}{2}{a}_y{t}^2$

we may write :

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

As t = 0, the value of t is $\frac{2v_0}{g}$.

Now form the equation,

$x=v_{0}t+\frac{1}{2}{a}_x{t}^2$

we can write,

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

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

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

substituting the value of $v_0$ in the above equation,

we get:

$l=8hsin\theta$

Therefore, the ball again hits the plain at a distance,

$l=8hsin\theta$Let the ball strikes the inclined plane at origin with velocity $v_0=\sqrt{2gh}$.

form the equation

$y=v_{oy}t+\frac{1}{2}{a}_y{t}^2$

we may write :

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

As t = 0, the value of t is $\frac{2v_0}{g}$.

Now form the equation,

$x=v_{0}t+\frac{1}{2}{a}_x{t}^2$

we can write,

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

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

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

substituting the value of $v_0$ in the above equation,

we get:

$l=8hsin\theta$

Therefore, the ball again hits the plain at a distance,

$l=8hsin\theta$