A cannon shell is fired to hit a target at a horizontal distance $R.$ However it breaks into two equal parts at its highest point. One part $A$ returns to the canon. The other part  

Just after the explosion , velocity of A becomes u cos $\theta as it retraces the path of original shell.$

By the principle of conservation of momentum ,velocity of the other part = 3u cos $\theta .$

The other part hits the ground after time t = $\sqrt{\frac{2H}{g}}=\sqrt{\frac{2}{g}\times \frac{u^2{\sin }^2\alpha }{2g}}=\frac{u \sin \alpha }{g} = \frac{T}{2}$

So the other part  will hit the ground at a horizontal distance ( measured from the point of explosion )= 3u cos$\theta$$\times \frac{\mathrm{u} \sin \mathrm{\theta }}{2}=\frac{3R}{2}$