a
(a)Let ball is projected vertically downward with velocity \(v\) from height \(h\)
Total energy at point \(A = \frac{1}{2}m{v^2} + mgh\)
During collision loss of energy is \(50\%\) and the ball rises up to same height. It means it possess only potential energy at same level.
\(50\%\) \(\left( {\frac{1}{2}m{v^2} + mgh} \right) = mgh\)
\(\frac{1}{2}\left( {\frac{1}{2}m{v^2} + mgh} \right) = mgh\)
\(v = \sqrt {2gh} = \sqrt {2 \times 10 \times 20} \)
\(v = 20\,m/s\)