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A body is projected vertically up with certain velocity. At a point ‘P’ in its path, the ratio of its potential to kinetic energies is 9 : 16. The ratio of velocity of projection to velocity at ‘P’ is

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25 : 16

5 : 4

9 : 25

3 : 4

answer is B.

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Detailed Solution

$\frac{{PE}_{h}}{{KE}_{h}} = \frac{36}{64} \Rightarrow \frac{mgh}{\frac{1}{2} {mv}^{2}} = \frac{36}{64}$

$\Rightarrow v^{2} = \frac{32}{9} gh$

Also $v^{2} = u^{2} - 2 gh (∵ gh = \frac{9}{32} v^{2})$

$\Rightarrow v^{2} = u^{2} - \frac{9}{16} v^{2} \Rightarrow \frac{u^{2}}{v^{2}} = \frac{25}{16}$

$\Rightarrow \frac{u}{v} = \frac{5}{4}$

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