A bullet fired from a gun falls at a distance 2 3 of its maximum range. The angle of projection of the bullet can be:

sin 2 θ = 3 /2 gives 2 θ = 6 0 ∘ , θ = 3 0 ∘ . Option B.

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PhysicsKinematics2-D & 3DMedium2 minQB

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A bullet fired from a gun falls at a distance $\frac{3}{2}$ of its maximum range. The angle of projection of the bullet can be:

Answer:

Solution:

$sin 2 θ = 3 /2$ gives $2 θ = 6 0^{\circ}$ , $θ = 3 0^{\circ}$ . Option B.

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:2-D & 3D

⏱ 2mℹ️ Source: QB

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