Question

The muzzle velocity of a projectile is 450 m/sec and the distance of the target is 16 km. Find the angle of elevation of the gun

Answer 1

$$R=\frac{v^2\sin{2\theta}}{g}\\16000=\frac{450^2\sin{2\theta}}{9.8}\\\theta=0.5\arcsin{\frac{16000\cdot9.8}{450^2}}\\\theta=25.4\degree$$