Question

A fighter plane flying horizontally at an altitude of $1.5km$ with speed $720km/h$ passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed $600m{s}^{-1}$ to hit the plane ? At what minimum altitude should the pilot fly the plane to avoid being hit ? ($Takeg=10m{s}^{-2}$)

Answer 1

Height of the fighter plane $=1.5km=1500$ m

Speed of the fighter plane, $v=720km/h=200$ m/s

Let be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.

Muzzle velocity of the gun, $u=600$ m/s

Time taken by the shell to hit the plane $=t$

Horizontal distance travelled by the shell $=u_{x}t$

Distance travelled by the plane $=vt$

The shell hits the plane. Hence, these two distances must be equal.

$u_{x}t=vt$

$uSin\theta =v$

$Sin\theta =v/u$

$=200/600=1/3=0.33$

$\theta =Si{n}^{-1}\left(0.33\right)=19.50$

In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell for any angle of launch.

$H_{max}=u^{2}si{n}^2(90-\theta )/2g={600}^2/(2\times 10)=16km$