Question

Single Correct Answer Type
A particle is projected at angle $\alpha$ to horizontal from foot of an inclined plane of angle ${30}^{\circ }$ with a particular velocity. If particle strikes the plane normally then $\alpha$ is equal to -

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

Answer 1

Answer: (A)

${30}^{\circ }+ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$t_{AB}=$ time of flight of projectile $=\frac{2u\sin (\alpha -{30}^{\circ })}{g\cos {30}^{\circ }}$
Now component of velocity along the plane becomes zero at point $B$.
$0=u\cos (\alpha -{30}^{\circ })-g\sin {30}^{\circ }\times T$
or $u\cos (\alpha -{30}^{\circ })=g\sin {30}^{\circ }\times \frac{2u\sin (\alpha -{30}^{\circ })}{g\cos {30}^{\circ }}$
or $\tan (\alpha -{30}^{\circ })=\frac{\cot {30}^{\circ }}{2}=\frac{\sqrt{3}}{2}$
or $\alpha ={30}^{\circ }+{\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$.