A projectile of mass 3 m is projected from ground with velocity $20\sqrt{2}m/s$ at $45°$. At highest point it explodes into two pieces. One of mass 2 m and the other of mass m. Mass 2m falls at a distance of 100m from point of protection. Find the distance of second mass from point of projection where it strikes the ground.$\left(g=10m{s}^{-2}\right)$

Range of the projectile in the absence of explosion

$\mathrm{R}=\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}}=\frac{{\left(20\sqrt{2}\right)}^2\sin 90°}{10}=80\mathrm{m}$

The path of center of mass of projectile will not change, i.e, ${\mathrm{x}}_{\mathrm{C}.\mathrm{O}.\mathrm{M}}$ is still 80m. Now, 
from the definition of center of mass.

${\mathrm{x}}_{\mathrm{COM}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}$

$80=\frac{\left(\mathrm{m}\right)\left({\mathrm{x}}_1\right)+\left(2\mathrm{m}\right)\left(100\right)}{\mathrm{m}+2\mathrm{m}}$

on solving this we get ${\mathrm{x}}_1=40\mathrm{m}$

Therefore, the mass m will fall at a distance ${\mathrm{x}}_1=40 \mathrm{cm}$ from point of projection.