JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2025
- Easy
- MCQ
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).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(24\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Range }=3 \mathrm{H}_{\max } \\ & \frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{~g}}=\frac{3 \mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\end{aligned}\)
\(2 \sin \theta \cos \theta=\frac{3}{2} \sin ^2 \theta\)
\(\tan \theta=\frac{4}{3} \Rightarrow \theta=53^{\circ}\)
\(\mathrm{R}=\frac{\mathrm{u}^2\left(2 \times \frac{3}{5} \times \frac{4}{5}\right)}{\mathrm{g}} \Rightarrow \frac{24 \mathrm{u}^2}{25 \mathrm{~g}}\)
\(2 \sin \theta \cos \theta=\frac{3}{2} \sin ^2 \theta\)
\(\tan \theta=\frac{4}{3} \Rightarrow \theta=53^{\circ}\)
\(\mathrm{R}=\frac{\mathrm{u}^2\left(2 \times \frac{3}{5} \times \frac{4}{5}\right)}{\mathrm{g}} \Rightarrow \frac{24 \mathrm{u}^2}{25 \mathrm{~g}}\)
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