A fighter plane flying horizontally at an altitude of 1.5 km with speed of 720 km/h passes directly overhead an anticraft gun. At what angle from the vertical should the gun be fixed for the shell with muzzle speed 400 m/s to hit the plane? 

Given: Let the fighter plane be flying horizontally with a speed u′ at height OP = 1.5 km, and point O represents the position of the anti-aircraft gun.

Let the shell be fired with velocity u making an angle θ with the vertical direction so that it hits the fighter plane at point B.

Resolving velocity u into components:

Vertical component = u cosθ
Horizontal component = u sinθ

If t is the time taken by the shell to hit the fighter plane, then the horizontal distance travelled by the fighter plane in time t with velocity u′ is equal to the horizontal distance travelled by the shell in time t with velocity u sinθ.

Therefore,

u′t = u sinθ · t ⇒ sinθ = u′ / u

Given:
u′ = 720 km/h = 720 × (1000 m) × (60 × 60)⁻¹ = 200 m/s
u = 400 m/s

∴ sinθ = 200 / 400 = 1/2

Hence, θ = 30° with the vertical.