Motion in a Plane question for JEE Main (JEE Main 2024 (8 Apr, Shift 2)), with solution

Answer: (A) $\tan^{-1}(4)$

$$\frac{u^2\sin2\theta}{g} = \frac{u^2\sin^2\theta}{2g} \;\Rightarrow\; 2\sin\theta\cos\theta = \frac{\sin^2\theta}{2} \;\Rightarrow\; \tan\theta = 4$$
So $\theta = \tan^{-1}(4)$.

Solution by Sreeraj P, M.Sc Physics