A ball of mass $2\,kg$ and another of mass $4\,kg$ are dropped together from a $60$ feet tall building. After a fall of $30$ feet each towards earth, their respective kinetic energies will be in the ratio of
A ball of mass $2\,kg$ and another of mass $4\,kg$ are dropped together from a $60$ feet tall building. After a fall of $30$ feet each towards earth, their respective kinetic energies will be in the ratio of
$\sqrt 2 :1$
$1:4$
$1:2$
$1:\sqrt 2 $
A ball of mass $2\,kg$ and another of mass $4\,kg$ are dropped together from a $60$ feet tall building. After a fall of $30$ feet each towards earth, their respective kinetic energies will be in the ratio of
Kinetic energy$ = \frac{1}{2}m{v^2}$
As both balls are falling through same height therefore they possess same velocity.
but $KE \propto m$ (If $v =$ constant)
$\frac{{{{\left( {KE} \right)}_1}}}{{{{\left( {KE} \right)}_2}}} = \frac{{{m_1}}}{{{m_2}}} = \frac{2}{4} = \frac{1}{2}$
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