Question

It is given that the plane has the inclination angle, $\theta ={45}^{\circ },$ From the height (h) of 5 m, a ball falls on the inclined plane. Then, the restitution coefficient is $e=3/4$. Find the length $AB$ where ball fall after collision.

Answer 1

$h=5m,\theta ={45}^o,e=\left(3/4\right)$

Here the velocity with which it would strike $=v=\sqrt{2g\times 5}=10m/sec$

After collision, let it make an angle $\beta$ with horizontal. the horizontal component of velocity $10\cos {45}^o$ will remain uncharged and the the velocity in the perpendicular direction to the plane after wllisine.

$\Rightarrow V_y=e\times 10\sin {45}^o$

$=\left(3/4\right)\times 10\times \frac{1}{\sqrt{2}}=\left(3.75\right)\sqrt{2}m/sec$

$V_x=10\cos {45}^o=5\sqrt{2}m/sec$

So, $u=\sqrt{V_x^2+V_y^2}=\sqrt{50+28.125}=sqrt78.125=8.83m/sec$

Angle of reflection from the wall $\beta ={\tan }^{-1}\left(\frac{3.75\sqrt{2}}{5\sqrt{2}}\right)={\tan }^{-1}\left(\frac{3}{4}\right)={37}^o$

$\Rightarrow$ Angle of projection $\alpha =90-(\theta +\beta )=90-({45}^o+{37}^o)=8^o$

Let the distance where it falls $=L$

$\Rightarrow x=L\cos \theta ,y=-L\sin \theta$

Angle of projection $\left(\alpha \right)=-8^o$

Using equation of trojectory, $y=x\tan \alpha -\frac{g{x}^2{\sec }^2\alpha }{2u^2}$

$\Rightarrow -\ell \sin \theta =\ell \cos \theta \times \tan 8^o-\frac{g}{2}\times \frac{\ell {\cos }^2\theta {\sec }^2{8}^o}{u^2}$

$\Rightarrow -\sin {45}^o=\cos {45}^o-\tan 8^o-\frac{10{\cos }^2{45}^o\sec 8^o}{(8.83{)}^2}\left(\ell \right)$

Solving the above equation we get,

$\ell =18.5m$.