Kinematics

JEE Main level, single correct, medium

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Question 242 of 308

A ball is projected from the ground with a speed 15ms−115{ms}^{-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

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Answer: (D) 44

The range and the maximum height of the projectile are given as below,
R=u2sin⁡2θg=2u2sin⁡θcos⁡θgH=u2sin⁡2θ2gR=\dfrac{{u}^{2}\sin 2\theta }{g}=\dfrac{2{u}^{2}\sin \theta \cos \theta }{g} H=\dfrac{{u}^{2}{\sin }^{2}\theta }{2g}
Therefore,
As H=RH=R
⇒tan⁡θ=4\Rightarrow \tan \theta =4

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