Question

A fighter plane flying horizontally at an altitude of $1.5 \,km$ with speed $720 \,km h^{-1}$ passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed $600 \,m s^{-1}$ to hit the plane. (Take $g = 10\, m s^{-2}$)

• A. $sin^{-1} \left(\frac{1}{3}\right)$
• B. $sin^{-1} \left(\frac{2}{3}\right)$
• C. $cos^{-1} \left(\frac{1}{3}\right)$
• D. $cos^{-1} \left(\frac{2}{3}\right)$

Answer 1

Answer: (A)

Here,
Speed of the plane, $v = 720 \,km\,h^{-1}$
$=720\times \frac{5}{18} ms^{-1} =200\, ms^{-1}$
Speed of the shell, $u = 600 \,ms^{-1}$
Let the shell will hit the plane at $L$ after time $t$ if fired at an angle $\theta$ with the vertical from $O$.
For hitting, horizontal distance travelled by the plane = horizontal distance travelled by the shell.
i,e., $v\times t=u\,sin \,\theta \times t$
$sin\, \theta=\frac{v}{u}=\frac{200\, ms^{-1}}{600\,ms^{-1}} =\frac{1}{3}$
$\theta =sin^{-1} (\frac{1}{3})$