A smooth inclined plane is inclined at an ang | vedclass

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(C) The body slides down a smooth inclined plane of length $l$ and height $h$,inclined at an angle $	heta$.The acceleration of the body along the inc

Vedclass · Accepted Answer

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

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(C) The body slides down a smooth inclined plane of length $l$ and height $h$,inclined at an angle $	heta$.The acceleration of the body along the incline is $a = g \sin 	heta$.The initial velocity $u = 0$.Using the second equation of motion,$S = ut + \frac{1}{2}at^2$,where $S = l$ and $u = 0$:$l = 0 + \frac{1}{2}(g \sin 	heta)t^2$$t^2 = \frac{2l}{g \sin 	heta}$From the geometry of the inclined plane,$\sin 	heta = \frac{h}{l}$,which implies $l = \frac{h}{\sin 	heta}$.Substituting $l$ in the expression for $t^2$:$t^2 = \frac{2(h / \sin 	heta)}{g \sin 	heta} = \frac{2h}{g \sin^2 	heta}$Taking the square root,we get $t = \frac{1}{\sin 	heta} \sqrt{\frac{2h}{g}}$.

$A$ smooth inclined plane is inclined at an angle $\theta$ with the horizontal. $A$ body starts from rest and slides down the inclined surface. The time taken by it to reach the bottom is

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