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(D) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.The maximum height $H$ is given by

Vedclass · Accepted Answer

$24$

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(D) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.According to the problem,$R = 3H$.Substituting the expressions,we get $\frac{2u^2 \sin 	heta \cos 	heta}{g} = 3 \left( \frac{u^2 \sin^2 	heta}{2g} ight)$.Simplifying this,$2 \sin 	heta \cos 	heta = \frac{3}{2} \sin^2 	heta$,which implies $	an 	heta = \frac{4}{3}$.Using $	an 	heta = \frac{4}{3}$,we find $\sin 	heta = \frac{4}{5}$ and $\cos 	heta = \frac{3}{5}$.Now,substitute these into the range formula: $R = \frac{u^2 (2 \sin 	heta \cos 	heta)}{g} = \frac{u^2 (2 	imes \frac{4}{5} 	imes \frac{3}{5})}{g} = \frac{24 u^2}{25 g}$.Comparing this with $\frac{n u^2}{25 g}$,we get $n = 24$.

$A$ 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 the value of $n$ is (Given $g$ is the acceleration due to gravity).

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