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If we throw a body upwards with velocity of $4 \mathrm{~m} / \mathrm{s}$, at what height does its kinetic energy reduce to half of the initial value?
(Take $g=10 \mathrm{~ms}^{-2}$ )
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(Take $g=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$0.4 m$
Key Idea : At a given height the half of the kinetic energy of the body is equal to its potential energy. Initial kinetic energy of the body
$\begin{aligned} & =\frac{1}{2} m v^2 \\ & =\frac{1}{2} m(4)^2=8 m\end{aligned}$
Let at height $h$, the kinetic energy reduces to half, ie, it becomes 4 m . It is also equal to potential energy. Hence,
$m g h=4 \mathrm{~m}$
or $h=\frac{4}{g}=\frac{4}{10}=0.4 \mathrm{~m}$
$\begin{aligned} & =\frac{1}{2} m v^2 \\ & =\frac{1}{2} m(4)^2=8 m\end{aligned}$
Let at height $h$, the kinetic energy reduces to half, ie, it becomes 4 m . It is also equal to potential energy. Hence,
$m g h=4 \mathrm{~m}$
or $h=\frac{4}{g}=\frac{4}{10}=0.4 \mathrm{~m}$
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