If we throw a body upwards with velocity of $4 m s^{- 1}$ at what height its kinetic energy reduces to half its initial value ? Take $g = 10 m / s^{2}$

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0.4 m

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Solution

The correct option is D
0.4 m

We know kinetic energy $K = \frac{1}{2} m v^{2} ∴ v \propto \sqrt{K}$

When kinetic energy of the body reduces to half its velocity becomes $v = \frac{u}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2} m / s$

From the equation $v^{2} = u^{2} - 2 g h \Rightarrow (2 \sqrt{2})^{2} = (4)^{2} - 2 \times 10 h ∴ h = \frac{16 - 8}{20} = 0.4 m$

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