An enemy plane is flying horizontally at an altitude of 2 km with a speed of 300 $m{s}^{-1}.$ An armyman with an anti-aircraft gun on the ground sights the enemy plane when it is directly overhead and fires a shell with a muzzle speed of 600 $m{s}^{-1}$. At what angle with the vertical should the gun be fired so as to hit the plane?

The correct option is A

${30}^{\circ }$

Let G be the position of the gun and E of the enemy plane flying horizontally with speed $u=300m{s}^{-1}$, when the shell is fired with a speed $v_0=600m{s}^{-1}$ in a direction $\theta$ with the horizontal . The horizontal component of $v_0$ is
$v_x=v_{0}cos\theta$
Let the shell hit the plane at point P and let t be the time taken for the shell to hit the plane. The horizontal distance EP travelled by the shell is same as the horizontal distance travelled by the plane.
$u\times t=v_x\times t$
$\Rightarrow u=v_x=v_{0}cos\theta$
$\Rightarrow cos\theta =\frac{u}{v_0}=\frac{300}{600}=0.5$
$\Rightarrow \theta ={60}^{\circ }$.
Angle with the vertical $={90}^{\circ }-\theta ={30}^{\circ }$.
Hence, the correct choice is (a).