Question

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

• A. $si{n}^{-1}\left(\frac{1}{3}\right)$
• B. $si{n}^{-1}\left(\frac{2}{3}\right)$
• C. $co{s}^{-1}\left(\frac{1}{3}\right)$
• D. $co{s}^{-1}\left(\frac{2}{3}\right)$

Answer 1

Answer: (A)

Here,

Speed of the plane, $v=720km{h}^{-1}$

$=720\times \frac{5}{18}m{s}^{-1}=200m{s}^{-1}$

Speed of the shell, $u=600m{s}^{-1}$

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., $v\times t=usin\theta \times t$
$sin\theta =\frac{v}{u}=\frac{200m{s}^{-1}}{600m{s}^{-1}}=\frac{1}{3}$

$\theta =si{n}^{-1}(\frac{1}{3})$