An object of mass 10 kg is launched from the ground at t = 0, at an angle of 37^o above the horizontal with a speed of 30 m/s. At some time after its launch, an explosion splits the projectile into two pieces. One piece of mass 4 kg is observed at (105 m,43 m) at t = 2 s. Find the location of second piece at t = 2 s?

As only the gravity force is acting on the system, the centre of mass of the system follows a parabolic path. 

$\begin{array}{l}\mathrm{At} \mathrm{t}=2\mathrm{s},\\ {\mathrm{x}}_{\mathrm{CM}}=30\cos {37}^{\circ }\times 2=48\mathrm{m}\\ {\mathrm{y}}_{\mathrm{CM}}=30\sin {37}^{\circ }\times 2-\frac{1}{2}\times 10\times 2^2=16\mathrm{m}\end{array}$

Let coordinates of the second piece, i.e., 6 kg piece be (x, y).Then 

  $\begin{array}{l}{\mathrm{x}}_{\mathrm{CM}}=48=\frac{6\mathrm{x}+4\times 105}{10}\Rightarrow \mathrm{x}=10\mathrm{m}\\ {\mathrm{y}}_{\mathrm{CM}}=16=\frac{6\mathrm{y}+4\times 43}{10}\Rightarrow \mathrm{y}=-2\mathrm{m}\end{array}$

Negative value of y shows that the second piece collides the ground before t = 2s.