question_answer

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) \[{{30}^{{}^\circ }}+{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]

B) \[{{45}^{0}}\]

C) \[{{60}^{0}}\]                       

D) \[{{30}^{0}}+{{\tan }^{-1}}(2\sqrt{3})\]

Correct Answer: A

Solution :

[a] \[{{t}_{AB}}\]=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. \[0=u\cos (\alpha -30{}^\circ )-gsin30{}^\circ \times T\] Or \[u\cos (\alpha -30{}^\circ )-gsin30{}^\circ \times \frac{2u\sin (\alpha -30{}^\circ )}{g\cos 30{}^\circ }\] Or \[\tan (\alpha -30{}^\circ )=\frac{\cot 30{}^\circ }{2}=\frac{\sqrt{3}}{2}\] Or \[\alpha =30{}^\circ +{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]