A 9.0 kg shell is fired at 50 m/s at 55° above level ground.

Step 1. Range. $R = \dfrac{u^2\sin 2\theta}{g} = \dfrac{2500\ \text{m}^2/\text{s}^2 \times \sin 110^\circ}{9.8\ \text{m/s}^2} = 239.7\ \text{m}$.

Step 2. The light piece. It falls from the top, at $\dfrac{R}{2} = 119.9\ \text{m}$.

Step 3. The heavy piece. $9.0\ \text{kg} \times 239.7\ \text{m} = 3.0\ \text{kg} \times 119.9\ \text{m} + 6.0\ \text{kg}\,x$, so $x = 299.6\ \text{m}$.