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 is 540 m/s, the firing angle θ for the bullet to hit the aircraft should be

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 is 540 m/s, the firing angle θ for the bullet to hit the aircraft should be <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ac8374d150101e254bdf51dafd796ec9-.png" /> (A) 73° (B) 30° (C) 60° (D) 45°

Tardigrade Admin · Accepted Answer

Displacement of aircraft in time t = horizontal displacement of projectile ⇒ 972× (5/18)t =V0 cos θ ⋅ t ⇒ cos θ = (54 × 5/540) =(1/2) ⇒ θ=60°

Q. 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 is $540 m / s$ , the firing angle $θ$ for the bullet to hit the aircraft should be

$7 3^{\circ}$

$3 0^{\circ}$

$6 0^{\circ}$

$4 5^{\circ}$

Displacement of aircraft in time t = horizontal displacement of projectile
$\Rightarrow 972 \times \frac{5}{18} t = V_{0} cos θ \cdot t$
$\Rightarrow cos θ = \frac{54 \times 5}{540} = \frac{1}{2}$
$\Rightarrow θ = 6 0^{\circ}$

Q. An aircraft moving with a speed of 1000km/h is at a height of 6000m, just overhead of an anti-aircraft gun. If the muzzle velocity of the gun is 540m/s, the firing angle θ for the bullet to hit the aircraft should be

73∘

30∘

60∘

45∘