Q 11-03-114JEE MainJEE Main 2022 (25 Jul, Shift 2)Easy
A ball is projected from the ground with a speed $15\ \text{m s}^{-1}$ at an angle $\theta$ with the horizontal so that its range and maximum height are equal. Then $\tan\theta$ will be equal to
Answer: (D) 4
$$\frac{2u^2\sin\theta\cos\theta}{g} = \frac{u^2\sin^2\theta}{2g}\ \Rightarrow\ 4\cos\theta = \sin\theta\ \Rightarrow\ \tan\theta = 4$$
Solution by Sreeraj P, M.Sc Physics