Question

A fighter plane flying horizontally at a altitude of 1.5 km with a speed of 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 a muzzle velocity of 600 m/s to hit the plane? At what minimum altitude should the pilot fly to avoid being hit? [Take g = 10 m /s²]

Answer 1

Given: h = 1.5 km = 1500 m,

u = 600 m/s,

y = 720 km/h = 720 × \(\frac{5}{18}\) = 200 m/s

To find: 

(i) Angle of firing (θ)

(ii) Minimum altitude (H)

Formula: H = \(\frac{u^2\,sin^2\,\theta}{2g}\)

Calculation:

Let the shell hit the plane t seconds after firing,

∴ 600 cos(90 – θ) × t = 200 t

∴ cos(90 – θ) = \(\frac{1}{6}\)

∴ 90° – θ = cos^-1 (\(\frac{1}{3}\))

cos^-1 (\(\frac{1}{3}\)) = 90° – θ

∴ 70°28’ = 90° – θ

∴ θ = 90° – cos^-1 (\(\frac{1}{3}\))

∴ θ = 19°47’ with vertical

To avoid being hit, the plane must be at a minimum height, i.e., maximum height reached by the shell.

From formula,

∴ H = 15.9 km

(i) Angle made by gun with the vertical is 19°47′.

(ii) Minimum altitude at which the pilot should fly is 15.9 km.