A body of mass m is projected at an angle $\mathrm{\theta }$ to the horizontal, the projectile at the highest point breaks into two fragments of equal masses. One of the fragments retraces its path to the point of projection. The velocity of the other fragment just after explosion is

Initial momentum at the highest point will be $\mathrm{mvcos\theta }$

And after explosion the velocity of the particle that retraces the path will be $\mathrm{vcos\theta }$ but in the opposite direction of the initial velocity.

As there is no external force acting on the body in horizontal direction, the momentum is conserved.

$\mathrm{mvcos\theta }=\frac{\mathrm{m}}{2}(-\mathrm{vcos\theta })+\frac{\mathrm{m}}{2}\left(\mathrm{v}\text{'}\right)\Rightarrow \mathrm{v}\text{'}=3\mathrm{vcos\theta }$