Gujarati
Hindi
5.Work, Energy, Power and Collision
hard

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

A

$4\sqrt 2 $

B

$1.5$

C

$\sqrt 2 $

D

None

Solution

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

$\frac{v^{2}}{v^{2} \cos ^{2} \theta}=9 \Rightarrow \cos \theta=\frac{1}{3}$

$\frac{\mathrm{R}}{\mathrm{H}}=4 \cot \theta=\sqrt{2}$

Standard 11
Physics

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