A shell of mass $2m$ fired with a speed $u$ at an angle $\theta$ to the horizontal explodes at the highest point of its trajectory into two fragments of mass $m$ each. If one fragment falls vertically, the distance at which the other fragment falls from the gun is given by

At the highest point of trajectory, the projectile has only a horizontal velocity which is $u \cos \theta$. After explosion, the fragment falling downwards has no horizontal velocity. If $u\text{'}$ is the horizontal velocity of the other fragment, the law of conservation of momentum gives 

   $\left(2m\right) u \cos \theta = m \times 0 + mu\text{'}$

which gives $u\text{'} = 2u \cos \theta$

Now, the time taken to reach the highest point (as well as the time taken to fall down from this point) is $\frac{u \sin \theta }{g}$. Therefore, the horizontal distance travelled by the other fragment is

          $$\begin{gathered} u \cos \theta \times \frac{u \sin \theta }{g}+2 u \cos \theta \times \frac{u \sin \theta }{g} \\ =\frac{u^2\sin 2\theta }{2g}+\frac{u^2\sin 2\theta }{g}=\frac{3u^2\sin 2\theta }{2g} \end{gathered}$$