A fighter plane is flying horizontally at an altitude of 1.5km with speed 720km/h. At what angle of sight (with respect to horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target.
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Solution
Formula used:T=√2Hg R=ux×T
Solution:
Given, H=1.5km=1500m ux=vo=720km/h=720×518=200m/s
Time of flight: T=√2Hg
Horizontal range (R) to hit the target (R=voT=vo√2Hg
From right triangle ABC tanϕ=HR=Hvo√2Hg tanϕ=1vo(√gH2)
Putting the values tanϕ=1200(√9.8×15002) ϕ=1200(√9.8×15002) ϕ=23∘