Question

A body is slipping from an inclined plane of height $h$ and length $l$. If the angle of inclination is $\theta$, the time taken by the body to come from the top to the bottom of this inclined plane is

• A. $\sqrt{\frac{2h}{g}}$
• B. $\sqrt{\frac{2l}{g}}$
• C. $\frac{1}{\sin\theta}\sqrt{\frac{2h}{g}}$
• D. $\sin\theta\sqrt{\frac{2h}{g}}$

Answer 1

Answer: (C)

$\frac{1}{\sin\theta}\sqrt{\frac{2h}{g}}$

Answer 2

Force down the plane $= mg\sin\theta$

$\therefore$ Acceleration down the plane $= g\sin\theta$

Since $l = 0 + \frac{1}{2}g\sin\theta t^{2}$

$\therefore$ $t^{2} = \frac{2l}{g\sin\theta} = \frac{2h}{g\sin^{2}\theta} \Rightarrow t = \frac{1}{\sin\theta}\sqrt{\frac{2h}{g}}$