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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 2 :3 . The height above the ground at which the ratio of PE to KE becomes 3 :2 is

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$\frac{2}{5} h$

$\frac{2}{3} h$

$\frac{3}{2} h$

$\frac{5}{2} h$

answer is A.

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Detailed Solution

$\frac{P E}{K E} = \frac{2}{3} \Rightarrow \frac{h}{H - h} = \frac{2}{3} \Rightarrow 3 h = 2 H - 2 h \Rightarrow 5 h = 2 H$

$\frac{P E}{K E} = \frac{x}{H - x} = \frac{3}{2} \Rightarrow \frac{x}{\frac{5}{2} h - x} = \frac{3}{2}$

$\Rightarrow 2 x = \frac{15 h}{2} - 3 x$

$\Rightarrow 5 x = \frac{1 5^{3} h}{2} \Rightarrow x = \frac{3}{2} h$

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