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JEE Main 2022 · 25 Jul · Shift 2, Physics question 48

A ball is projected from the ground with a speed 15 ms−1^{-1} at an angle θ\theta with horizontal so that its range and maximum height are equal, then 'tan⁡θ\tan\theta' will be equal to :

Choose one option
Answer and worked solution

Official answer: D

Range = height. 2sin⁡θcos⁡θg=sin⁡2θ2g\dfrac{2\sin\theta\cos\theta}g=\dfrac{\sin^2\theta}{2g}, so tan⁡θ=4\tan\theta=4.

(D)

Topics:Projectile motion
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