First slide

Projectile motion

Question

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

Moderate

A

73^o
B

30^o
C

60^o
D

45^o

Solution

Displacement of aircraft in time t = horizontal displacement of projectile
$\begin{array}{l} \Rightarrow 972 \times \frac{5}{18} t = V_{0} \cos θ \cdot t \\ \Rightarrow \cos θ = \frac{54 \times 5}{540} = \frac{1}{2} \Rightarrow θ = 60^{\circ} \end{array}$

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