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).

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Range }=3 \mathrm{H}_{\max } \ & \frac{\mathrm{u}^2 \sin 2 	heta}{\mathrm{~g}}=\frac{3 \mathrm{u}^2 \sin ^2 	heta}{2 \mathrm{~g}}\end{aligned}$ $2 \sin 	heta \cos 	heta=\frac{3}{2} \sin ^2 	heta$ $	an 	heta=\frac{4}{3} \Rightarrow 	heta=53^{\circ}$ $\mathrm{R}=\frac{\mathrm{u}^2\left(2 	imes \frac{3}{5} 	imes \frac{4}{5}ight)}{\mathrm{g}} \Rightarrow \frac{24 \mathrm{u}^2}{25 \mathrm{~g}}$

A particle is projected with velocity $u$ so that its horizontal range is three times the maximum height…

Solution