Q 11-03-175JEE MainJEE Main 2025 (3 Apr, Shift 2)Easy
A particle is projected with velocity $u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $\dfrac{nu^2}{25g}$, where the value of $n$ is: (Given $g$ is the acceleration due to gravity)
Answer: (D) $24$
$$R = 3H \Rightarrow \frac{2u^2\sin\theta\cos\theta}{g} = 3\cdot\frac{u^2\sin^2\theta}{2g} \Rightarrow \tan\theta = \frac43$$
So $\sin\theta = \frac45$, $\cos\theta = \frac35$:
$$R = \frac{2u^2}{g}\cdot\frac45\cdot\frac35 = \frac{24u^2}{25g} \Rightarrow n = 24$$
Solution by Sreeraj P, M.Sc Physics