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Question

$\mathrm{K}_{\max }=\frac{1}{2} \mathrm{mv}^{2}, \mathrm{K}_{\min }=\frac{1}{2} \mathrm{mv}^{2} \cos ^{2} 	heta$

$\frac{v^{2}}{v^{2} \cos ^{2} 	h

Vedclass · Accepted Answer

$\sqrt 2 $

Vedclass · Answer

$\mathrm{K}_{\max }=\frac{1}{2} \mathrm{mv}^{2}, \mathrm{K}_{\min }=\frac{1}{2} \mathrm{mv}^{2} \cos ^{2} 	heta$

$\frac{v^{2}}{v^{2} \cos ^{2} 	heta}=9 \Rightarrow \cos 	heta=\frac{1}{3}$

$\frac{\mathrm{R}}{\mathrm{H}}=4 \cot 	heta=\sqrt{2}$

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

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