Maximum range,
$R_{\max }=\frac{u^2 \sin 2 \theta}{g}=\frac{u^2}{g}=16 \mathrm{~km}$
Range is maximum when $\theta=45^{\circ}$ Initial momentum at the highest point
$=m u \cos 45^{\circ}=\frac{m u}{\sqrt{2}}$
After explosion, the projectile breaks into two equal masses. As one mass drops vertically downwards, hence its velocity and momentum is zero in the $x$-direction.
Hence, if $v$ be the velocity of second mass, then according to law of conservation of momentum,
$m u \cos 45^{\circ}=\frac{m}{2} v \quad \therefore \quad v=2 u \cos 45^{\circ}$
Horizontal distance covered from the time of explosion
$\begin{aligned} & =v \times \frac{T}{2}=v \times \frac{1}{2} \times \frac{2 u \sin 45^{\circ}}{g} \\ & =2 u \cos 45^{\circ} \times \frac{u \sin 45^{\circ}}{g} \\ & =\frac{u^2}{g} \times \sin 90^{\circ}=\frac{u^2}{g}=16 \mathrm{~km}\end{aligned}$