Q65 / 75 · JEE Main 2025 (3 Apr Shift 2) · Physics
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 $\frac{n u^2}{25 g}$, where value of $n$ is : (Given ' $g$ ' is the acceleration due to gravity).
- A$6$
- B$18$
- C$12$
- Correct answerD$24$
Solution
$\begin{aligned} & \text { Range }=3 \mathrm{H}_{\max } \\ & \frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{~g}}=\frac{3 \mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\end{aligned}$
$2 \sin \theta \cos \theta=\frac{3}{2} \sin ^2 \theta$
$\tan \theta=\frac{4}{3} \Rightarrow \theta=53^{\circ}$
$\mathrm{R}=\frac{\mathrm{u}^2\left(2 \times \frac{3}{5} \times \frac{4}{5}\right)}{\mathrm{g}} \Rightarrow \frac{24 \mathrm{u}^2}{25 \mathrm{~g}}$
$2 \sin \theta \cos \theta=\frac{3}{2} \sin ^2 \theta$
$\tan \theta=\frac{4}{3} \Rightarrow \theta=53^{\circ}$
$\mathrm{R}=\frac{\mathrm{u}^2\left(2 \times \frac{3}{5} \times \frac{4}{5}\right)}{\mathrm{g}} \Rightarrow \frac{24 \mathrm{u}^2}{25 \mathrm{~g}}$
Hence, the correct answer is option: 'D'.