Kinematics

JEE Main level, single correct, medium

207 of 308

Question 207 of 308

A particle is projected with velocity uu so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as nu225g\frac{n u^2}{25 g}, where value of nn is : (Given ' gg ' is the acceleration due to gravity).

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Answer: (D) 2424

 Range =3Hmax⁡u2sin⁡2θ g=3u2sin⁡2θ2 g\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}
2sin⁡θcos⁡θ=32sin⁡2θ2 \sin \theta \cos \theta=\frac{3}{2} \sin ^2 \theta
tan⁡θ=43⇒θ=53∘\tan \theta=\frac{4}{3} \Rightarrow \theta=53^{\circ}
R=u2(2×35×45)g⇒24u225 g\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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