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(B) Let the mass of the body be $m$ and the initial velocity be $u$. At maximum height $h$, the total energy is $E = mgh = \frac{1}{2}mu^2$.At a heigh

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$3$:$2$

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(B) Let the mass of the body be $m$ and the initial velocity be $u$. At maximum height $h$, the total energy is $E = mgh = \frac{1}{2}mu^2$.At a height $y = 0.4h$ from the ground:The potential energy is $PE = mgy = mg(0.4h) = 0.4mgh$.According to the law of conservation of energy, the total energy remains constant.$KE + PE = E$$KE + 0.4mgh = mgh$$KE = mgh - 0.4mgh = 0.6mgh$.The ratio of kinetic energy to potential energy is:$\frac{KE}{PE} = \frac{0.6mgh}{0.4mgh} = \frac{0.6}{0.4} = \frac{6}{4} = 3:2$.

$A$ body thrown vertically upwards from the ground reaches a maximum height $h$. The ratio of the kinetic and potential energies of the body at a height $40 \%$ of $h$ from the ground is

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