Solve the following :
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?

Solution:

After falling down, ball will have a projectile motion.

$\alpha=\frac{\pi}{2}-2 \theta$

$u=\sqrt{2 g h}, O P=l$

$x=l \cos \theta \quad y=-l \sin \theta$

Equation of trajectory

$\Rightarrow y=x \tan \alpha-\frac{g x^{2} \sec ^{2} \alpha}{2 u^{2}}$

$\Rightarrow-l \sin \theta=l \cos \theta \cdot \tan \left(\frac{\pi}{2}-20\right)-\frac{g l^{2} \sec ^{2}\left(\frac{\pi}{2}-2 \theta\right)}{2(2 g h)} \cos ^{2} \theta$

$\Rightarrow l=\frac{4 h}{\cos ^{2} \theta \csc ^{2} \theta}(\sin \theta+\cos \theta \cot 2 \theta)$

$=16 h \sin ^{2} \theta \times \frac{\cos \theta}{2 \sin \theta \cos \theta}=8 h \sin \theta$