Answer
$\frac{-{{v}_{0}}+\sqrt{{{v}_{0}}^{2}+2gs}}{g}$
Work Step by Step
$s={{v}_{0}}t+\frac{g{{t}^{2}}}{2}$
Solving the equation for $t$,
Multiply by $2$ on both sides of the equation.
$\begin{align}
& s={{v}_{0}}t+\frac{g{{t}^{2}}}{2} \\
& s\times 2=\left( {{v}_{0}}t+\frac{g{{t}^{2}}}{2} \right)\times 2 \\
& 2s=2{{v}_{0}}t+g{{t}^{2}}
\end{align}$
Then, subtract $2s$ on both sides of the equation,
$\begin{align}
& 2s=2{{v}_{0}}t+g{{t}^{2}} \\
& 2s-2s=2{{v}_{0}}t+g{{t}^{2}}-2s \\
& 0=2{{v}_{0}}t+g{{t}^{2}}-2s
\end{align}$
$g{{t}^{2}}+2{{v}_{0}}t-2s=0$
Now, using the quadratic formula,
$t=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substitute the values of the equation in the quadratic formula,
$a=g,b=2{{v}_{{}^\circ }},c=-2s$
Then, the equation is,
$\begin{align}
& t=\frac{-2{{v}_{0}}\pm \sqrt{{{\left( 2{{v}_{0}} \right)}^{2}}-4\times g\times \left( -2s \right)}}{2\times g} \\
& =\frac{-2{{v}_{0}}\pm \sqrt{4{{v}_{0}}^{2}+8gs}}{2g} \\
& =\frac{-{{v}_{0}}\pm \sqrt{{{v}_{0}}^{2}+2gs}}{g}
\end{align}$
We only consider positive time values:
$t=\frac{-{{v}_{0}}+\sqrt{{{v}_{0}}^{2}+2gs}}{g}$
Thus, the value is $t=\frac{-{{v}_{0}}+\sqrt{{{v}_{0}}^{2}+2gs}}{g}$.
