A fighter plane flying horizontal at an altitude of 1.5 Km with speed of 720 Km $h^{-1}$ passes directly overhead an anticarft gun. At what angle from the vertical should the gun be fired from the shell with muzzle speed 400 $m{s}^{-1}$ to hit the plane in shortest time.?

The correct option is B 14.47
consider the aircraft gun is launched at an angle $\theta$ from vertical at a speed of $u=400m/s$. it follows a projectile motion

consider they collide at a time t.

in time $t$,fighter plane travels.

$d=vt$

also for aircraft gun

$u\sin \theta t=d$

that gives

$vt=u\sin thetat$

$\Rightarrow u\sin \theta =v\Rightarrow \sin \theta =\frac{v}{u}=\frac{720\times \frac{5}{18}}{400}=\frac{4\times 5}{40}=\frac{1}{4}$

$\Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{4}\right)\simeq 14.470$

so at $\theta =14.470$,its unit the plane in minimum time