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Question

A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/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=1.5 km=1500 m
ux=vo=720km/h=720×518=200m/s

image
Time of flight:
T=2Hg
Horizontal range (R) to hit the target
(R=voT=vo2Hg
From right triangle ABC
tanϕ=HR=Hvo2Hg
tanϕ=1vo(gH2)
Putting the values
tanϕ=1200(9.8×15002)
ϕ=1200(9.8×15002)
ϕ=23

Final Answer:
ϕ=(HR)=tan1(1vogH2)=23

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