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 ms^(-1)` to hit the plane ? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take `g = 10 ms^(-2)`).

Explanation

In Fig. O be the position of gun and A be the position of plane. The speed of the plane,
`v = (720 xx 1000)/(60 xx 60) = 200 ms^(-1)`
The speed of the shell, `u = 600 m//s`
Let the shell will hit the plane at B after time t if fired at an angle `theta` with the vertical from O then the horizontal distance travelled by shel in time t is the same as the distance covered by plane.
i.e. `u_(x) xx t = v t` or `u sin theta t = v t`
or `sin theta = (v)/(u) = (200)/(600) = 0.3333 = sin 19.5^(@)`. or `theta = 19.5^(@)`

The plane will not be hit by the bullet from the gun if it is flying at a minimum height which is maximum height (H) attained by bullet after firing from gun.
Here `H = (u^(2) sin^(2) (90 - theta))/(2g) = (u^(2) cos^(2) theta)/(g)`
`= ((600)^(2) xx (cos 19.5)^(2))/(2 xx 10)`
`= 16000 m`
= 16 km