A) \[20\,m{{s}^{-1}}\]
B) \[15\,m{{s}^{-1}}\]
C) \[10\,m{{s}^{-1}}\]
D) \[5\,m{{s}^{-1}}\]
Correct Answer: A
Solution :
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\]You need to login to perform this action.
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