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When a body is projected vertically up from ground, the ratio of its potential to kinetic energies at a height 'h' above the ground is 3 : 5. The height above ground at which the ratio of potential to kinetic energies becomes 5 : 3 is

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$\frac{5}{3}$ h

$\frac{4}{7}$ h

$\frac{3}{7}$ h

$\frac{7}{4}$ h

answer is A.

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Detailed Solution

$\frac{P E}{K E} = \frac{3}{5} = \frac{h}{H - h}$

$\frac{H}{h} = \frac{8}{3}$

$\frac{P E}{K E} = \frac{5}{3} = \frac{h^{1}}{H - h^{1}} \Rightarrow \frac{H}{h^{1}} = \frac{8}{5}$

$\frac{h^{1}}{h} = \frac{5}{3} \Rightarrow h^{1} = \frac{5}{3} h$

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