## Algebra 2 (1st Edition)

$\frac{v_0^2}{g}$
Plugging in $h_0=0$ into the formula we get: $\frac{v_0\sqrt{(v_0)^2+2gh_0}}{g}=\frac{v_0\sqrt{(v_0)^2+2g(0)}}{g}=\frac{v_0\sqrt{(v_0)^2+0}}{g}=\frac{v_0\sqrt{(v_0)^2}}{g}=\frac{v_0(v_0)}{g}=\frac{v_0^2}{g}$