JEE Mains · Physics · STD 11 - 3.2 motion in plane
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 \(12\)
- D \(24\)
Answer & Solution
Correct Answer
(D) \(24\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Range }=3 \mathrm{H}_{\max } \\ & \frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{~g}}=\frac{3 \mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\end{aligned}\) \(2 \sin \theta \cos \theta=\frac{3}{2} \sin ^2 \theta\)…
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