Question

A particle is projected with certain velocity at an angle $\alpha$ above the horizontal from the foot of an inclined plane having inclination of ${30}^{\circ }$. If the particle strikes the plane normally then $\alpha$ is:

• A. $\alpha ={30}^{\circ }+{\tan }^{-1}\frac{2}{\sqrt{3}}$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. $\alpha ={30}^{\circ }+{\tan }^{-1}\frac{\sqrt{3}}{2}$

Answer 1

The correct option is D $\alpha ={30}^{\circ }+{\tan }^{-1}\frac{\sqrt{3}}{2}$

If the projectile hits the inclined plane normally, it means that component of final velocity ($v$) along the incline is zero i.e $v_x=0$
Also,
$u_x=u\cos (\alpha -{30}^{\circ })$
$u_y=u\sin (\alpha -{30}^{\circ })$
$a_x=-g\sin \theta =-g\sin {30}^{\circ }$, $a_y=-g\cos \theta =-g\cos {30}^{\circ }$

Resolving the whole motion as $2$ separate independent $1$-D motion along $x$ and $y$ direction respectively, and applying kinematic equation:

$\therefore v_x=u_x+a_{x}t$

$\Rightarrow 0=u\cos (\alpha -{30}^{\circ })-g\sin {30}^{\circ }t$
$\Rightarrow t=\frac{u\cos (\alpha -{30}^{\circ })}{g\sin {30}^{\circ }}$......$i$

Now, time of flight for a projectile on inclined plane is given by the formula:

$\Rightarrow T=\frac{2u_y}{|a_y|}=\frac{2u\sin (\alpha -{30}^{\circ })}{g\cos {30}^{\circ }}$..........$ii$

Equating two expressions ($i$) and ($ii$) as both represent the time of flight for particle projected on the inclined plane, we get:

$\therefore T=t$

$\Rightarrow \frac{2u\sin (\alpha -{30}^{\circ })}{g\cos {30}^{\circ }}=\frac{u\cos (\alpha -{30}^{\circ })}{g\sin {30}^{\circ }}$
$\Rightarrow \tan (\alpha -{30}^{\circ })=\frac{1}{2\tan {30}^{\circ }}=\frac{\sqrt{3}}{2}$

$\therefore \alpha -{30}^{\circ }={\tan }^{-1}\frac{\sqrt{3}}{2}$
Hence, $\alpha ={30}^{\circ }+{\tan }^{-1}\frac{\sqrt{3}}{2}$