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A body of mass $2 \mathrm{~kg}$ is thrown up vertically with K.E. of 490 joules. If the acceleration due to gravity is $9.8 \mathrm{~m} / \mathrm{s}^2$, then the height at which the K.E. of the body becomes half its original value is given by
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$12.5 \mathrm{~m}$
Let $h$ is that height at which the kinetic energy of the body becomes half its original value i.e. half of its kinetic energy will convert into potential energy
$\therefore m g h=\frac{490}{2} \Rightarrow 2 \times 9.8 \times h=\frac{490}{2} \Rightarrow h=12.5 m \text {. }$
$\therefore m g h=\frac{490}{2} \Rightarrow 2 \times 9.8 \times h=\frac{490}{2} \Rightarrow h=12.5 m \text {. }$
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