Kinematics Question 525
Question: An aircraft moving with a speed of 1000 km/h is at a height of 6000 m, just overhead of an anti-aircraft gun. If the muzzle velocity of the gun it’s 540 m/s, the firing angle $ \theta $ for the bullet to hit the aircraft should be
Options:
A) $ 73{}^\circ $
B) $ 30{}^\circ $
C) $ 60{}^\circ $
D) $ 45{}^\circ $
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Answer:
Correct Answer: C
Solution:
[c] Displacement of aircraft in time t=horizontal displacement of projectile
$ \Rightarrow 972\times \frac{5}{18}t=V _{0}\cos \theta .t $
$ \Rightarrow \cos \theta =\frac{54\times 5}{540}=\frac{1}{2}\Rightarrow \theta =60{}^\circ $
