First slide

Projection Under uniform Acceleration

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 ${ms}^{- 2}$ )

Moderate

A

$\tan^{- 1} (\frac{\sqrt{3}}{4})$
B

$\tan^{- 1} (\frac{\sqrt{3}}{2})$
C

$\tan^{- 1} (\frac{1}{2})$
D

$\tan^{- 1} (2)$

Solution

Here, u = 720 km $h^{- 1}$
= $720 \times \frac{5}{18} {ms}^{- 1} = 200 {ms}^{- 1}$
H = 1.5km = 1.5 $\times 1000 m = 1500 m$
Time taken by the bomb to attack the target
t = $\sqrt{\frac{2 H}{g}}$ = $\sqrt{\frac{2 \times 1500 m}{10 {ms}^{- 2}}} = \sqrt{300} s = 10 \sqrt{3} s$
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) $α = \tan^{- 1} (\frac{H}{R})$
$α = \tan^{- 1} (\frac{1500}{2000 \sqrt{3}}) = \tan^{- 1} (\frac{3}{4 \sqrt{3}}) = \tan^{- 1} (\frac{\sqrt{3}}{4})$

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