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  4. A body of mass 5 kg is thrown vertically up with a kinetic energy of 490 J. The height at which the kinetic energy of the body becomes half of the original value is (acceleration due to gravity = 9.8 ms-2)

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Q. A body of mass $5\,kg$ is thrown vertically up with a kinetic energy of $490\,J$. The height at which the kinetic energy of the body becomes half of the original value is (acceleration due to gravity = $9.8\, ms^{-2}$)

KCETKCET 2011Work, Energy and Power

Solution:

Given, $m=5\, kg$
and $KE =490\, J$
By the law of conservation of energy
$\frac{1}{2} m u^{2} =\frac{1}{2} m v^{2}+m g h$
$490 =245+5 \times 9.8 \times h$
$h =\frac{490-245}{5 \times 9.8}$
$h =\frac{245}{49}$
$=5\, m$