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Question

An aircraft moving with a speed of 972 km/hr 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, then the firing angle θ with the horizontal for the bullet to hit the aircraft should be 


  1. 73
  2. 30
  3. 45
  4. 60


Solution

The correct option is D 60
Velocity of aircraft in horizontal direction
v=972 km/hr=972×518=270 m/s
Muzzle velocity of gun is
v=540 m/s
The component of velocity of bullet in horizontal direction will be
vx=vcosθ=540cosθ

In time interval t s, the horizontal distance covered by bullet will be
D=vxt=(540cosθ)t ...(i)
Horizontal distance covered by aircraft in same time interval t s:
D=270t ...(ii)

For bullet to hit the aircraft, the condition should be:
D=D
or (540cosθ)t=270t
cosθ=270540=12
θ=60

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