Question

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

• A. ${30}^{\circ }+{\tan }^{-1}\left(\frac{1}{2\sqrt{3}}\right)$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${30}^{\circ }+{\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer 1

The correct option is D ${30}^{\circ }+{\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Let $v$ be the final velocity of the particle when it strikes the inclined plane.
As the final velocity is perpendicular to the inclined plane so,
$v_x=0$
Using first equation of motion along $x$-axis, we get
$v_x=u_x-g\sin {30}^{\circ }T$ where $T$ is the time of flight.
$v_x=0\Rightarrow T=\frac{u\cos (a-{30}^{\circ })}{g\sin {30}^{\circ }}..\left(i\right)$
Using second equation of motion along $y$-axis, we get
$s_y=u_{y}T-\frac{1}{2}g\cos {30}^{\circ }{T}^2$
As displacement along $y$-direction is zero for the projectile motion
$S_y=0$
$\Rightarrow 0=u\sin (\alpha -{30}^{\circ })T-\frac{1}{2}g\cos {30}^{\circ }{T}^2$
$\Rightarrow u\sin (\alpha -{30}^{\circ })T=\frac{1}{2}g\cos {30}^{\circ }{T}^2$
$\Rightarrow T=\frac{2u\sin (\alpha -{30}^{\circ })}{g\cos {30}^{\circ }}...\left(ii\right)$

On comparing $\left(i\right)$ and $\left(ii\right)$, we get
$\frac{2u\sin (a-{30}^{\circ })}{g\cos {30}^{\circ }}=\frac{u\cos (a-{30}^{\circ })}{g\sin {30}^{\circ }}\Rightarrow \tan (a-{30}^{\circ })=\frac{\sqrt{3}}{2}\Rightarrow a={30}^{\circ }+{\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$