question_answer

A ball is projected vertically down with an initial velocity from a height of 20 m onto a horizontal floor. During the impact it loses 50% of its energy and rebounds to the same height. The initial velocity of its projection is [EAMCET (Engg.) 2000]

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\]