A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight when the target is seen, should the pilot drop the bomb in order to attack the target?
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’