Kinematics Question 516

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

Options:

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}) $

Show Answer

Answer:

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}}( \frac{\sqrt{3}}{2} ) $



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