Question

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

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

Answer 1

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

Velocity of aircraft in horizontal direction
$v=972\text{km/hr}=972\times \frac{5}{18}=270\text{m/s}$
Muzzle velocity of gun is
$v^{\prime }=540\text{m/s}$
The component of velocity of bullet in horizontal direction will be
$v_x^{\prime }=v^{\prime }\cos \theta =540\cos \theta$

In time interval $t\text{s}$, the horizontal distance covered by bullet will be
$D^{\prime }=v_x^{\prime }t=\left(540\cos \theta \right)t...\left(i\right)$
Horizontal distance covered by aircraft in same time interval ${}^{\prime }{t}^{\prime }\text{s}$:
$D=270t...\left(ii\right)$

For bullet to hit the aircraft, the condition should be:
$D^{\prime }=D$
or $\left(540\cos \theta \right)t=270t$
$\Rightarrow \cos \theta =\frac{270}{540}=\frac{1}{2}$
$\therefore \theta ={60}^{\circ }$