Kinematics Question 189
Question: A body is slipping from an inclined plane of height $ h $ and length $ l $ . If the angle of inclination it’s $ \theta $ , the time taken by the body to come from the top to the bottom of this inclined plane is
Options:
A) $ \sqrt{\frac{2h}{g}} $
B) $ \sqrt{\frac{2l}{g}} $
C) $ \frac{1}{\sin \theta }\sqrt{\frac{2h}{g}} $
D) $ \sin \theta \sqrt{\frac{2h}{g}} $
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Answer:
Correct Answer: C
Solution:
Force down the plane $ =mg\sin \theta $
$ \therefore $ Acceleration down the plane $ =g\sin \theta $ Since $ l=0+\frac{1}{2}g\sin \theta t^{2} $
$ \therefore $ $ t^{2}=\frac{2l}{g\sin \theta }=\frac{2h}{g{{\sin }^{2}}\theta }\Rightarrow t=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}} $
