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Question

Physics Question on projectile motion

The horizontal range of a projectile is 434\sqrt{3} times its maximum height, then the angle of projection is

A

30∘30^{\circ}

B

45∘45^{\circ}

C

60∘60^{\circ}

D

90∘90^{\circ}

Answer

30∘30^{\circ}

Explanation

Solution

Using the relation R=u2sin⁡2θgR=\frac{u^{2} \sin 2 \theta}{g} and H=u2sin⁡2θ2gH=\frac{u^{2} \sin ^{2} \theta}{2 g} Given u2sin⁡2θg=43(u2sin⁡2θ2g)\frac{u^{2} \sin 2 \theta}{g}=4 \sqrt{3}\left(\frac{u^{2} \sin ^{2} \theta}{2 g}\right) 2sin⁡θcos⁡θ=43sin⁡2θ22 \sin \theta \cos \theta=\frac{4 \sqrt{3} \sin ^{2} \theta}{2} 13=sin⁡θcos⁡θ=tan⁡θ\frac{1}{\sqrt{3}}=\frac{\sin \theta}{\cos \theta}=\tan \theta or tan⁡θ=13\tan \theta=\frac{1}{\sqrt{3}} Hence, θ=30∘\theta=30^{\circ}