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 $g = 10ms^{- 2},\tan 23^{o} = 0.43$)

• A. $23^{o}$
• B. $32^{o}$
• C. $12^{o}$
• D. $42^{o}$

Answer 1

Answer: (A)

$23^{o}$

Answer 2

Here, $u = 720kmh^{- 1}$

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

H = $1.5km = 1.5 \times 1000m = 1500m$

Time taken by the bomb to attack the target

$t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2 \times 1500m}{10ms^{- 1}}} = \sqrt{300}s = 10\sqrt{3}s$

$R = u \times t = 200ms^{- 1} \times 10\sqrt{3}s = 2000\sqrt{3}m$

From Figure, Angle of sight (w.r.t. horizontal)

$\alpha = \tan^{- 1}\left( \frac{H}{T} \right)$

$\alpha = \tan^{- 1}\left( \frac{1500}{2000\sqrt{3}} \right) = \tan^{- 1}\left( \frac{3}{4\sqrt{3}} \right) = \tan^{- 1}\left( \frac{\sqrt{3}}{4} \right)\alpha = \tan^{- 1}(0.43)or\alpha = 23{^\circ}$