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 = `720 * 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 = `ut + 1/2 gt^2`

That gives, `1500 = 0 + 1/2 10t^2`

`t = sqrt(300) = 10sqrt(3) s`

Now distance covered by the plane is `ut = 200 xx 10sqrt(3) = 2000sqrt(3)`

Hence, `tan theta = 1500/(2000sqrt(3)) = sqrt(3)/4`

Therefore, `θ = tan^-1  sqrt(3)/4` = 23°42'