Question

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).

Chapter Motion In A Plane Page 88

Answer 1

Speed of fighter plane =720km/h
= 720 × 5/18 = 200 m/s
Height of fighter plane = 1.5km
= 1500 m

Muzzle speed of the shell = 600m/s .

Let the shell inclined at angle ∅ with vertical as shown in figure .
Hence,
Vertical component of muzzle speed if the shell = ucos∅.
And horizontal component of it = usin∅ .
Now,
Horizontal distance covered by plane = horizontal distance covered by shell
Speed of plane × t = usin∅× t
200 = 600sin∅
Sin∅ = 1/3
∅ = sin-¹(1/3)

e.g gun fired at sin-¹(1/3) with vertical to hit the fighter plane .

Again,
Speed of Fighter plane inclined (90-∅) with horizontal .for avoiding the hit of shell speed of fighter plane (U)= speed of shell = 600m/s
Hence,
maximum height = Usin²(90-∅)/2g where , U is speed of fighter plane .

H = (600)² × cos²∅/2g
= 16000 m
= 16 km
Hence, minimum attitude should 16km for avoiding the hit of shell.