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,

\(\begin{array}{l}
Gain\,in\,kinetic\,energy\, = loss\,in\,potential\\
energy\,i.e.\,\frac{1}{2}m{v^2} - \frac{1}{2}mv_0^2 = mgh\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {where\,m\,is\,the\,mass\,of\,the\,ball} \right)\\
or\,\,\,{v^2} - v_0^2 = 2gh\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\\
Now,\,when\,the\,ball\,collides\,with\,the\,\\
ground,\,50\% \,of\,its\,energy\,is\,lost\,and\\
it\,rebounds\,to\,the\,same\,height\,h.
\end{array}\)

\(\begin{array}{l}
\therefore \,\,\,\,\,\frac{{50}}{{100}}\left( {\frac{1}{2}m{v^2}} \right) = mgh\\
\frac{1}{4}{v^2} = gh\,\,or\,\,{v^2} = 4gh\\
Subsitiuding\,this\,value\,of\,{v^2}\,in\,eqn.\\
\left( i \right),\,we\,get\\
\,\,\,\,\,\,\,\,4\,gh\, - v_0^2 = 2gh\\
or\,\,\,\,\,\,v_0^2\, = 4gh - 2gh - 2gh\,\,or\,\,{v_0} = \sqrt {2gh} 
\end{array}\)

\(\begin{array}{l}
Here,\,g\, = 10\,m{s^{ - 2}}\,and\,h = 20\,m\\
\therefore \,\,\,\,{v_0} = \sqrt {2\left( {10\,m{s^{ - 2}}} \right)\left( {20m} \right)}  = 20m{s^{ - 1}}
\end{array}\)