#### Question

A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/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 600 m s^{–1} to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take *g *= 10 m s^{–}^{2}).

#### Solution 1

Height of the fighter plane = 1.5 km = 1500 m

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

Let *θ *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*

*`u sin theta = v`*

`sin theta = v/u`

= `200/600 = 1/3 = 0.33`

`theta = sin^(-1)(0.33)`

`= 19.5^@`

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.

`:.H = (u^2sin^2(90-theta))/"2g"`

`=((600)^2cos^2theta)/(2g)`

`= (360000xxcos^2 19.5)/(2xx10)`

` =18000xx(0.943)^2`

=16006.482 m

=16 km

#### Solution 2

Velocity of plane, v_{p}=720 x 5/180 ms^{-1}=200 ms^{-1}

Velocity of shell = 600 `ms^(-1)`

`sin theta = 200/600 = 1/3`

or `theta = sin^(-1)(1/3) = 19.47^@`

This angle is with the verticle.

Let h be the required minimum height

Using equation

`v^2 - u^2 = 2as` we get

`(0)^2-(600 costheta)^2 = -2xx10 xxh`

or `h = (600xx600(1-sin^2theta))/20`

=`30xx600(1-1/9) = 8/9xx30xx600m` = 16 km