Question
A ball is thrown vertically downwards from a height of $$20\,m$$ with an initial velocity $${v_0}.$$ It collides with the ground, loses $$50\% $$ of its energy in collision and rebounds to the same height. The initial velocity $${v_0}$$ is (Take, $$g = 10\,m{s^{ - 2}}$$ )
A.
$$14\,m{s^{ - 1}}$$
B.
$$20\,m{s^{ - 1}}$$
C.
$$28\,m{s^{ - 1}}$$
D.
$$10\,m{s^{ - 1}}$$
Answer :
$$20\,m{s^{ - 1}}$$
Solution :
Suppose a ball rebounds with speed $$v,$$
$$\eqalign{
& v = \sqrt {2gh} = \sqrt {2 \times 10 \times 20} \cr
& = 20\,m/s \cr} $$
Energy of a ball just after rebound,
$$E = \frac{1}{2}m{v^2} = 200\,m$$
As, $$50\% $$ of energy loses in collision means just before collision energy is $$400\,m.$$
According to law of conservation of energy, we have
$$\eqalign{
& \frac{1}{2}mv_0^2 + mgh = 400\,m \cr
& \Rightarrow \frac{1}{2}mv_0^2 + m \times 10 \times 20 = 400\,m \cr
& \Rightarrow {v_0} = 20\,m/s \cr} $$