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 $600\,ms^{-1}$ to hit the plane.
(Take $g = 10\,ms^{-2}$)

• 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, $v=750\,km\,h^{-1}$
$=720\times\frac{5}{18}ms^{-1}$
$=200\,ms^{-1}$
Speed of the shell, $u = 600\,ms^{-1}$
Let the shell 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, $v \times t = usin \theta \times t$
$sin\,\theta=\frac{v}{u}$
$=\frac{200\,ms^{-1}}{600\,ms^{-1}}=\frac{1}{3}$
$\theta=sin^{-1}\left(\frac{1}{3}\right)$