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  4. A ball of mass 0.5 kg is dropped from the height of 10 m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ....... m. (Use g=10 m / s 2).

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Q. A ball of mass $0.5\, kg$ is dropped from the height of $10\, m$. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ....... $m$. (Use $g=10\, m / s ^{2}$).

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Solution:

$v^{2}=u^{2}+2 a s$
$100=0+2(10) s$
$S=5\, m$
Height from ground $=10-5=5\, m$