એક બોલને 20;m ઊંચાઇએથી પ્રારંભિક v_0 | vedclass

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Question

The situation is shown in the figure.

Let $v$ be  the velocity of the ball with which it collides with ground. Then according to the law of co

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$20$

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The situation is shown in the figure.

Let $v$ be  the velocity of the ball with which it collides with ground. Then according to the law of conservation of energy,

The situation is shown in the figure.

Let $v$ be  the velocity of the ball with which it collides with ground. Then according to the law of conservation of energy,

$\begin{array}{l}
Gain\,in\,kinetic\,energy\, = loss\,in\,potential\
energy\,i.e.\,\frac{1}{2}m{v^2} - \frac{1}{2}mv_0^2 = mgh\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {where\,m\,is\,the\,mass\,of\,the\,ball} ight)\
or\,\,\,{v^2} - v_0^2 = 2gh\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)\
Now,\,when\,the\,ball\,collides\,with\,the\,\
ground,\,50\% \,of\,its\,energy\,is\,lost\,and\
it\,rebounds\,to\,the\,same\,height\,h.
\end{array}$

$\begin{array}{l}
	herefore \,\,\,\,\,\frac{{50}}{{100}}\left( {\frac{1}{2}m{v^2}} ight) = mgh\
\frac{1}{4}{v^2} = gh\,\,or\,\,{v^2} = 4gh\
Subsitiuding\,this\,value\,of\,{v^2}\,in\,eqn.\
\left( i ight),\,we\,get\
\,\,\,\,\,\,\,\,4\,gh\, - v_0^2 = 2gh\
or\,\,\,\,\,\,v_0^2\, = 4gh - 2gh - 2gh\,\,or\,\,{v_0} = \sqrt {2gh} 
\end{array}$

$\begin{array}{l}
Here,\,g\, = 10\,m{s^{ - 2}}\,and\,h = 20\,m\
	herefore \,\,\,\,{v_0} = \sqrt {2\left( {10\,m{s^{ - 2}}} ight)\left( {20m} ight)}  = 20m{s^{ - 1}}
\end{array}$

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