Question

A fighter plane flying horizontally at an altitude of $2km$ with a speed of $540km/h$ passes directly overhead an anti-aircraft gun. If the gun can fire a bullet at the muzzle speed of $500m{s}^{-1}$, at what angle with the vertical should the gunfire so that the bullet hits the plane?

• A. ${\cos }^{-1}\left(\frac{3}{10}\right)$
• B. ${\sin }^{-1}\left(\frac{3}{10}\right)$
• C. ${\tan }^{-1}\left(\frac{3}{10}\right)$
• D. ${\tan }^{-1}\left(1\right)$

Answer 1

Answer: (B)

${\sin }^{-1}\left(\frac{3}{10}\right)$

for bullet the muzzle velocity $=500$ m/s$=500\times \frac{{10}^{-3}km}{\frac{1}{60\times 60}hr}=1800$ km/hr

Assume bullet fired at a angle $\theta$ and velocity of gun is zero

$u_h=1800\sin \theta$ km/hr
$u_v=1800\cos \theta$ km/hr

where $u_h$ and $u_v$ are horizontal and vertical velocities

Given: $S_v=2km$

Let time taken be $t$ hr

Dist. traveled by plane, $S=540\times t$ km

This is equal to range of bullet
$\therefore R=u_h\times t=540\times t$
$\Rightarrow 1800\sin \theta \times t=540\times t$

$\Rightarrow \sin \theta =\frac{3}{10}$

$\theta ={\sin }^{-1}\left(\frac{3}{10}\right)$