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STANDARDPhysics-Projectile Motion

STANDARD Physics Range and Maximum Height MCQ Question

Type: MCQ-numerical-Medium

The speed of a projectile at its maximum height is 32\frac{\sqrt{3}}{2} times its initial speed. If the range of the projectile is P times the maximum height attained by it, then P equals

A

43\frac{4}{3}

B

232\sqrt{3}

C

434\sqrt{3}

D

34\frac{3}{4}

Correct Answer

Option C

Detailed Explanation

Let uu be the initial speed and θ\theta is the angle of the projection. Speed at the maximum height, vH=ucos⁡θ=32uv_H = u \cos \theta = \frac{\sqrt{3}}{2} u. Therefore, cos⁡θ=32\cos \theta = \frac{\sqrt{3}}{2} or θ=30∘\theta = 30^\circ. Range, R=u2sin⁡2θgR = \frac{u^2 \sin 2\theta}{g} and maximum height, H=u2sin⁡2θ2gH = \frac{u^2 \sin^2 \theta}{2g}. As R=PHR = PH, P=4tan⁡30∘=43P = \frac{4}{\tan 30^\circ} = 4\sqrt{3}.

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