A body of mass $2 kg$ is thrown vertically up with kinetic energy of $500 J$ . What will be height at which the kinetic energy of the body becomes ${\frac{2}{3}}^{r d}$ of original value? $(g = 10 {m/s}^{2})$

6 m

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9.86 m

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8.33 m

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6.23 m

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Solution

The correct option is C $8.33 m$
Given $m = 2 kg$ and Kinetic energy $(K_{i}$ ) $= 500 J$
From the law of conservation of energy
$K_{i} + U_{i} = K_{f} + U_{f}$ [Here, sum of kinetic energy and potential energy will be constant ]
$\Rightarrow K_{i} + 0 = \frac{2 K_{i}}{3} + m g h$ (Given $K_{f} = \frac{2 K_{i}}{3}$ and as ground is taken as reference $U_{i} = 0$ )

$\Rightarrow \frac{K_{i}}{3} = m g h$
$\Rightarrow \frac{500}{3} = 2 \times 10 \times h$
$\Rightarrow h = 8.33 m$

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