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).

Answer 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_xt

Distance travelled by the plane = vt

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

u_xt = 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

Answer 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