A fighter plane is flying horizontally at an altitude

Given,

u = 720 km/h = 200 m/s

Let t be the time at which pilot drops the bomb, then Q will be the point vertically above the target T.

We also know that the horizontal velocity of the bomb is equal to the velocity of the fighter plane. But the vertical component is zero. When the bomb reaches TQ, it is a free falling object with initial velocity equal to zero.

u = 0

H = 1.5 km = 10 m/s2

H = ut + 1/2 gt2

Substituting the values, we get

t = 10√3 second.

Let the distance between PQ be ut.

Therefore, PQ = 2000√3 m

tan θ = √3/4

tan θ = tan-123o42’

θ = 23o42’