First slide

Newton's laws of motion

Question

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

Easy

A

$\sqrt{\frac{2 h}{g}}$
B

$\sqrt{\frac{2 l}{g}}$
C

$\frac{1}{\sin θ} \sqrt{\frac{2 h}{g}}$
D

$\sin θ \sqrt{\frac{2 h}{g}}$

Solution

$Force down the plane = m g \sin θ$
$Acceleration down the plane = g \sin θ$
$Since l = 0 + \frac{1}{2} g \sin θ t^{2}$
$t^{2} = \frac{2 l}{g \sin θ} = \frac{2 h}{g \sin^{2} θ} \Rightarrow t = \frac{1}{\sin θ} \sqrt{\frac{2 h}{g}}$

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