Question

An aircraft moving with a speed of $540\text{km/h}$ is at a height of $6000\text{m}$, just overhead of an anti aircraft gun. If the muzzle velocity of the bullet is $300\text{m/s}$, the firing angle $Q$ from the horizontal for the bullet to hit the aircraft should be

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${37}^{\circ }$

Answer 1

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

Let bullet hits the aircraft in time $t$.
$\therefore \text{Displacment of aircraft in time}t=\text{horizontal displacement of projectile}$
Given $v_{\text{aircraft}}=540\times \frac{5}{18}=150\text{m/s}$
Horizontal velocity of bullet
$(v_{\text{bullet}}{)}_x=V_0\times \cos Q=300\cos Q$
Since horizontal motion of bullet is non accelerated,
$150\times t=300\cos Q\times t$
$\Rightarrow \text{cos Q}=\frac{150}{300}=\frac{1}{2}$
$\therefore Q={60}^{\circ }$