A smooth inclined plane is inclined at an ang | vedclass

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Question

$\begin{array}{l}
So\,by\,{m{second}}\,equation\,of\,motion,\,we\,get\
S = ut + \frac{1}{2}a{t^2}\
here\,S = \ell ,u = 0,a = g\sin 	heta \
t

Vedclass · Accepted Answer

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

Vedclass · Answer

$\begin{array}{l}
So\,by\,{m{second}}\,equation\,of\,motion,\,we\,get\
S = ut + \frac{1}{2}a{t^2}\
here\,S = \ell ,u = 0,a = g\sin 	heta \
t = \sqrt {\frac{{2\ell }}{a}}  = \sqrt {\frac{{2h}}{{g{{\sin }^2}	heta }}}  = \frac{1}{{\sin 	heta }}\sqrt {\frac{{2h}}{g}} \
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sin 	heta  = \frac{h}{\ell }} ight)
\end{array}$

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

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