A shell fired at an angle $\text{'}\theta \text{'}$, velocity of the shell at the highest point is $u\cos \theta$

Let its mass be $m$

The speed of one piece is $u\cos \theta$ and let the speed of other piece be $v$.

Hence at the highest point, from conservation of momentum 

$mu\cos \theta = -\frac{m}{2}u\cos \theta +\frac{m}{2}\mathrm{v}$

$mu\cos \theta +\frac{m}{2} u\cos \theta =\frac{m}{2}\mathrm{v}$

$\frac{2mu\cos \theta +mu\cos \theta }{2}=\frac{m}{2}\mathrm{v}$

$m\mathrm{v}=3mu\cos \theta$

$\mathrm{v}=3u\cos \theta$ .......(i)

The kinetic energy

$E_1=\frac{1}{2}\times \frac{m}{2}{u}^2{\cos }^2\theta$

$E_1=\frac{1}{4}m{u}^2{\cos }^2\theta$ .......(ii)

Similarly,

$E_2=\frac{1}{2}\times \frac{m}{2}\times 9u^2{\cos }^2\theta$

$E_2=\frac{9}{4}m{u}^2{\cos }^2\theta$ ......(iii)

The relation between the $E_1$ and $E_2$

$E_2=9\times E_1$ ( from the equation (ii) )

$E_2=9E_1$