Kinematics

JEE Main level, single correct, medium

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Question 1 of 308

A projectile is launched with velocity uu at an angle θ=45∘\theta = 45^\circ with horizontal. The ratio of maximum height HH to horizontal range RR is:

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Answer: (D) 1/41/4

HR=u2sin⁡2θ/(2g)u2sin⁡2θ/g=tan⁡θ4=tan⁡45∘4=14\frac{H}{R} = \frac{u^2 \sin^2\theta / (2g)}{u^2 \sin 2\theta / g} = \frac{\tan\theta}{4} = \frac{\tan 45^\circ}{4} = \frac{1}{4}.

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