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(C) The force acting down the inclined plane is $F = mg \sin 	heta$. Therefore,the acceleration of the body down the plane is $a = g \sin 	heta$. Us

Vedclass · Accepted Answer

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

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(C) The force acting down the inclined plane is $F = mg \sin 	heta$. Therefore,the acceleration of the body down the plane is $a = g \sin 	heta$. Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $s = l$,$u = 0$,and $a = g \sin 	heta$: $l = 0 + \frac{1}{2}(g \sin 	heta)t^2$. From the geometry of the inclined plane,we know that $\sin 	heta = \frac{h}{l}$,which implies $l = \frac{h}{\sin 	heta}$. Substituting $l$ into the equation: $\frac{h}{\sin 	heta} = \frac{1}{2}g \sin 	heta t^2$. $t^2 = \frac{2h}{g \sin^2 	heta}$. Taking the square root on both sides,we get $t = \frac{1}{\sin 	heta} \sqrt{\frac{2h}{g}}$.

$A$ 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

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