• # question_answer A ball is thrown vertically downwards from a height of 20 m with an initial velocity$v_{0}^{2}.$It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity ${{v}_{0}}$is: (Take$g=10\,m{{s}^{-2}}$) A) $~10\,m{{s}^{-1}}$      B) $~14\,m{{s}^{-1}}$C) $~20\,m{{s}^{-1}}$      D) $~28\,m{{s}^{-1}}$

Let ball rebound with speed v, so $v=\sqrt{2gh}=\sqrt{2\times 10\times 20}=20\,m/s$ Energy just after rebound $E=\frac{1}{2}\times m\times {{v}^{2}}=200\,m$ 50% energy loses in collision means just before collision energy is 400 m By using energy conservation, $\frac{1}{2}mv_{0}^{2}+mgh=400\,m$ $\Rightarrow$$\frac{1}{2}mv_{0}^{2}+m\times 10\times 20=400\Rightarrow {{v}_{0}}=20\,m/s$