A body of mass 2 kg attains a maximum height ‘h’ when thrown vertically up. At what height, its kinetic energy is 75% of its initial value?

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$\frac{h}{4}$

$\frac{h}{2}$

$\frac{3 h}{4}$

$\frac{3 h}{8}$

answer is A.

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Detailed Solution

according law of conservation of mechanical energy
$(KE + PE)_{i} = (KE + PE)_{f}$
$\begin{aligned} \frac{1}{2} m u^{2} = \frac{3}{4} (\frac{1}{2} m u^{2}) + m g h \\ h = \frac{1}{4} (\frac{u^{2}}{2 g}) = \frac{1}{4} H_{max} \end{aligned}$

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