An aircraft moving with a speed of 972 km/h is at a height of 6000 m, just overhead of an anti-aircraft gun. If the muzzle velocity of the gun is 540 m/s, the firing angle θ for the bullet to hit the aircraft should be
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a
73o
b
30o
c
60o
d
45o
answer is C.
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Detailed Solution
Displacement of aircraft in time t = horizontal displacement of projectile⇒ 972×518t=V0cosθ⋅t⇒ cosθ=54×5540=12⇒θ=60∘