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) \[30{}^\circ +\,te{{n}^{-1}}(\frac{\sqrt{3}}{2})\]

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

C) \[60{}^\circ \]             

D) \[30{}^\circ \,+\,te{{n}^{-1}}(2\sqrt{3})\]

Correct Answer: A

Solution :

[a] \[{{t}_{AB}}\]=time of flight of projectile \[=\frac{2u\sin (\alpha -30{}^\circ )}{g\cos 30{}^\circ }\] Now the component of velocity along the plane becomes zero at point B. \[\theta =u\cos (\alpha -30{}^\circ )-gsin30{}^\circ \times {{t}_{AB}}\] Or \[u\cos (\alpha -30{}^\circ )=gsin30{}^\circ \times \frac{2u\sin (\alpha -30{}^\circ )}{g\cos 30{}^\circ }\] Or \[\tan (\alpha -30{}^\circ )=\frac{\cot 30{}^\circ }{2}=\frac{\sqrt{3}}{2}\] \[\alpha =30{}^\circ +{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]