Question

A fighter plane is flying horizontally at an altitude of $1.5 \,km$ with speed $720\, km h^{-1}$. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target? (Take $g = 10\, m s^{-2}$)

• A. $tan ^{-1}\left(\frac{\sqrt{3}}{4}\right)$
• B. $tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• C. $tan ^{-1}\left(\frac{\sqrt{1}}{2}\right)$
• D. $tan ^{-1}(2)$

Answer 1

Answer: (A)

Here, $u = 720\,km\,h^{-1}$
$= 720 \times \frac{5}{18} ms^{-1} = 200\,ms^{-1}$
$H = 1.5 \,km = 1.5 \times 1000 \,m = 1500\,m$
Time taken by the bomb to attack the target
$t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2 \times 1500\,m}{10\,ms^{-1}}} $
$ = \sqrt{300} s = 10 \sqrt{3} s$
$R = u \times t = 200\,ms^{-1} \times 10\sqrt{3} s $
$ = 2000\sqrt{3} m$
From figure,
Angle of sight (w. r. t horizontal) $\alpha = \tan^{-1} \left(\frac{H}{R}\right)$
$\alpha = \tan^{-1} \left(\frac{1500}{2000\sqrt{3}}\right) = \tan^{-1}\left(\frac{3}{4\sqrt{3}}\right)$
$ = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right)$