A projectile has a maximum range of $16 \mathrm{~km}$. At the highest point of its motion, it explodes into two equal masses. One mass drops vertically downwards. The horizontal distance covered by the other mass from the time of explosion is

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 $\begin{aligned} & =v \times \frac{T}{2}=v \times \frac{1}{2} \times \frac{2 u \sin 45^{\circ}}{g} \\ & =2 u \cos 45^{\circ} \times \frac{u \sin 45^{\circ}}{g} \\ & =\frac{u^2}{g} \times \sin 90^{\circ}=\frac{u^2}{g}=16 \mathrm{~km}\end{aligned}$