Kinematics Question 159
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}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Force down the plane = mg $ \sin \theta $ Acceleration down the plane = g $ \sin \theta $ Since l =0+ $ \frac{1}{2}g\sin \theta t^{2} $
$ l^{2}=\frac{2}{g\sin \theta }=\frac{2h}{g{{\sin }^{2}}\theta }\Rightarrow t=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}} $
