Question

A fighter plane flying horizontally at an attitude of 1.5 km with a speed of 720 km/h passes directly overhead of an antiaircraft gun. At what angle from the horizontal should the gun be fired for the shell muzzle speed 600 m/s to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit. [ g = 10 $m/s^2$]

Answer 1

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

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

Let $\theta$ 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, $u=600m/s$

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

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

Distance traveled by the plane $=vt$

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

therefore,

$u_{x}t=vt$

$u\sin \theta =v$

$\sin \theta =\frac{v}{u}$

$=\frac{200}{600}=\frac{1}{3}=0.33$

$\theta ={\sin }^{-1}\left(0.33\right)={19.5}^o$

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.

$\therefore H=\frac{u^2{\sin }^2(90-\theta )}{2g}$

$=\frac{(600{)}^2{\cos }^2\theta }{2g}$

$=\frac{360000\times {\cos }^2\theta }{2g}$

$=16006.482m$

$=16km$