A projectile is fired at a speed of $100m{s}^{-1}$ at an angle of ${37}^0$ above the horizontal. At the highest point, the projectile breaks into two parts of mass ratio 1 : 3, the smaller coming to rest. Find the distance from the launching point to the point where the heavier piece lands

At the highest point, the projectile has horizontal velocity. The lighter part comes to rest. Hence, the heavier part will move with increased horizontal velocity. In vertical direction, both parts have zero velocity and undergo same acceleration. Thus, both will hit the ground together. As internal forces do not affect the motion  of the centre of mass, the centre of mass hits the ground at the position where the original projectile would have landed. The range of the original projectile is.

$x_{cm}=\frac{2u^2\sin \theta \cos \theta }{g}=\frac{2\times {10}^4\times \frac{3}{5}\times \frac{4}{5}}{10}m=960m$

The centre of mass will hit the ground at this position. As the smaller block comes to rest after breaking it falls down vertically and hits the ground at half of the range, i.e, at $x=480m.$ If the heavier block hits the ground at $x_2,$ then

$x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$   $\Rightarrow 960=\frac{\left(m\right)\left(480\right)+\left(3m\right)\left(x_2\right)}{\left(m+3m\right)}$

$\Rightarrow x_2=1120m$