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} ) $
