For a projectile motion, the range R is 'n' t | vedclass

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

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$	an^{-1}(\frac{4}{n})$

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(C) The horizontal range $R$ of a projectile 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 $H = \frac{u^2 \sin^2 	heta}{2g}$.Given that $R = nH$, we substitute the expressions:$\frac{2u^2 \sin 	heta \cos 	heta}{g} = n \left( \frac{u^2 \sin^2 	heta}{2g} ight)$.Simplifying both sides:$2 \cos 	heta = \frac{n \sin 	heta}{2}$.Rearranging for $	an 	heta$:$\frac{\sin 	heta}{\cos 	heta} = \frac{4}{n}$.Therefore, $	an 	heta = \frac{4}{n}$, which implies $	heta = 	an^{-1}(\frac{4}{n})$.

For a projectile motion, the range $R$ is '$n$' times the maximum height $H$. So the angle of projection is

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