Question

A fighter jet is flying horizontally at a height of 1 km above ground at 200 m/s . at at the moment shown, it drops a bomb and the Canon shown on ground fires a shell from point A which destroys the bomb in air . Find the Angle of projection of shell ($\theta$) .
Given Speed of shell = 200 m/s ​

Answer 1

Answer:

An aircraft is flying horizontally with a constant velocity 200ms−1, at a height 1 km above the ground. At the moment shown, a bomb is released from the aircraft and the cannon gun below fires a shell with initial speed 200ms−1, at some angle θ.

Suppose the shell destroys the bomb at time t. Then for horizontal motion,

t(200+200cosθ)=3×1000

∴t(1+cosθ)=53              ...(i)

For vertical motion,

21gt2+(200sinθ)t−21gt2=1000

sinθt=5             ...(ii)

From (i) and (ii), 1+cosθsinθ=31

On solving we get θ=60o

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