### Q.30:- 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:-**

Height of the fighter plane = 1.5 km = 1500 m

Speed of the fighter plane,

Let

Speed of the fighter plane,

*v*= 720 km/h = 200 m/sLet

*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 θ = v

Sin θ = v / u

= 200 / 600 = 1/3 = 0.33

θ = Sin^{-1}(0.33) = 19.5^{0}

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^{2}Sin^{2} (90 – θ) **/** 2g

= (600)^{2} Cos^{2} θ / 2g

= 360000 X Cos^{2} 19.5 **/** 2 X 10

= 16006.482 m

≈ 16 km