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What is the equation when rearranged to solve for $v?\frac{1}{2}m{v}^2=mgh$

$v\frac{?}{\sqrt[\phantom{2}]{gh-\frac{1}{2}}}=$

$v=\frac{2}{\sqrt[\phantom{2}]{gh}}$

$v\frac{?}{\sqrt[\phantom{2}]{2gh}}=$

$1$