Question

An aircraft moving with a speed of $972km/h$ is at a height of $6000m$, just overhead of an anti-aircraft gun. If the muzzle velocity of the gun $540m/s$, the firing angle $\theta$ for the bullet to hit the aircraft should be

Diagram

• A. ${45}^{\circ }$
• B. ${73}^{\circ }$
• C. ${60}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

Given:
velocity of aircraft
$\left(v_a\right)=972km/h=972\times \frac{5}{18}=270m/s$
Height of aircraft $H=6000m$
Muzzle velocity of gun $\left(V_0\right)=540m/s$
Thus, from the given diagram ,
Dislacement of aircraft in time $t$
=horizontal displacement of projectile
$v_a\times t=V_0\cos \theta \cdot t270\times t=V_0\cos \theta \cdot t\cos \theta =\frac{54\times 5}{540}=\frac{1}{2}\theta ={60}^{\circ }$