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

Vedclass · Accepted Answer

$\frac{16}{n^2}$

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(B) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2	heta}{g}$ and the maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.Given $R = nH$,we have $\frac{u^2 (2 \sin 	heta \cos 	heta)}{g} = n \left[ \frac{u^2 \sin^2 	heta}{2g} ight]$.Simplifying this,$2 \cos 	heta = \frac{n \sin 	heta}{2}$,which gives $	an 	heta = \frac{4}{n}$.At maximum height,the vertical component of velocity is $0$,so the velocity is $v = u \cos 	heta$.The potential energy is $PE = mgH$ and the kinetic energy is $KE = \frac{1}{2} m v^2 = \frac{1}{2} m (u \cos 	heta)^2$.The ratio $\frac{PE}{KE} = \frac{mgH}{\frac{1}{2} m u^2 \cos^2 	heta} = \frac{2g \left( \frac{u^2 \sin^2 	heta}{2g} ight)}{u^2 \cos^2 	heta} = 	an^2 	heta$.Substituting $	an 	heta = \frac{4}{n}$,we get $\frac{PE}{KE} = \left( \frac{4}{n} ight)^2 = \frac{16}{n^2}$.

$A$ ball is projected in a manner such that its horizontal range is $n$ times the maximum height. Find the ratio of potential energy to kinetic energy at the maximum height.

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