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(C) The kinetic energy of a projectile is maximum at the point of projection and at the ground level (where speed is $v$),and minimum at the highest p

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$\sqrt{2}$

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(C) The kinetic energy of a projectile is maximum at the point of projection and at the ground level (where speed is $v$),and minimum at the highest point (where speed is $v \cos 	heta$).Given $\frac{K_{\max}}{K_{\min}} = 9$,we have $\frac{\frac{1}{2}mv^2}{\frac{1}{2}mv^2 \cos^2 	heta} = 9$.This simplifies to $\frac{1}{\cos^2 	heta} = 9$,which means $\cos^2 	heta = \frac{1}{9}$,so $\cos 	heta = \frac{1}{3}$.Since $\cos 	heta = \frac{1}{3}$,we find $\sin 	heta = \sqrt{1 - (\frac{1}{3})^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.Thus,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{2\sqrt{2}/3}{1/3} = 2\sqrt{2}$.The range $R$ is given by $R = \frac{v^2 \sin 2	heta}{g} = \frac{2v^2 \sin 	heta \cos 	heta}{g}$ and the maximum height $H$ is $H = \frac{v^2 \sin^2 	heta}{2g}$.The ratio $\frac{R}{H} = \frac{2v^2 \sin 	heta \cos 	heta / g}{v^2 \sin^2 	heta / 2g} = 4 \cot 	heta$.Substituting $\cot 	heta = \frac{1}{	an 	heta} = \frac{1}{2\sqrt{2}}$,we get $\frac{R}{H} = 4 	imes \frac{1}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

$A$ particle is projected from level ground. Its kinetic energy $K$ changes due to gravity such that $\frac{K_{\max}}{K_{\min}} = 9$. The ratio of the range $R$ to the maximum height $H$ attained during its flight is:

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