A ball is thrown vertically down from a height of 40m from the ground with an initial velocity ‘v’. The ball hits the ground, loses one-third of its total mechanical energy and rebounds back to the same height. If the acceleration due to gravity is 10ms^–2, the value of ‘v’ is

If the initial velocity is u,

After falling through a height h=40m

${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{gh}={\mathrm{u}}^2+800$

So just before collision the total mechanical energy is $\frac{1}{2}\mathrm{m}({\mathrm{u}}^2+800)$

As the mechanical energy is reduced to 2/3rd. 

And that will reach to a height of 40m.

$\frac{2}{3}\left[\frac{1}{2}\mathrm{m}({\mathrm{u}}^2+800)\right]=\mathrm{m}(10\times 40)$

$\frac{1}{3}\left({\mathrm{u}}^2+800\right)=400\Rightarrow {\mathrm{u}}^2=400$

$\mathrm{u}=20\mathrm{m}/\mathrm{s}$.