__Question 4.30__:

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}).

__Solution__:

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.

*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

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