Question

A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. 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?

Answer 1

The velocity of the fighter plane at a height h = 1.5 km = 1500 m is given as u = 720 km/h = ${\displaystyle 720\cdot \frac{5}{18}}$ = 200 m/s

Let the plane drops the bomb t seconds before the target is exactly below the plane.

Then in t seconds, the bomb must cover the vertical distance of 1500 m under free fall with an initial velocity of zero.

Hence, h = ${\displaystyle ut+\frac{1}{2}{>}^2}$

That gives, ${\displaystyle 1500=0+\frac{1}{2}10t^2}$

${\displaystyle t=\sqrt{300}=10\sqrt{3}s}$

Now distance covered by the plane is ${\displaystyle ut=200\times 10\sqrt{3}=2000\sqrt{3}}$

Hence, ${\displaystyle \tan \theta =\frac{1500}{2000\sqrt{3}}=\frac{\sqrt{3}}{4}}$

Therefore, ${\displaystyle \theta ={\tan }^{-1} \frac{\sqrt{3}}{4}}$ = 23°42'