Kinematics Question 518
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 }}+{{\tan }^{-1}}( \frac{\sqrt{3}}{2} ) $
B) $ 45^{0} $
C) $ 60^{0} $
D) $ 30^{0}+{{\tan }^{-1}}(2\sqrt{3}) $
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Answer:
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}}( \frac{\sqrt{3}}{2} ) $
