Answer
t $=$ $\frac{ v_{0}±\sqrt {(v_{0})^{2}+ 64(s_{0}-h)}}{32}$
Work Step by Step
We need to solve the quadratic equation $h$ $=$ $-$$16$$t^{2}$ $+$ $v_{0}$$t$ $+$ $s_{0}$.
Shift the $h$ from Left Hand Side (LHS) to Right Hand Side (RHS), so that the equation becomes,
$-$$16$$t^{2}$ $+$ $v_{0}$$t$ $+$ $s_{0}$ $-$$h$ $=$ $0$
We will use the quadratic formula to solve this equation.
The quadratic formula is as follows,
$\frac{ -b ±\sqrt {b^{2}- 4ac}}{2a}$
Here $a$$=$ $-$16
$b$ $=$ $v_{0}$
$c$ $=$ $s_{0}$$-$$h$
Putting the following in quadratic formula, we get
$\frac{ -(v_{0}) ±\sqrt {(v_{0})^{2}- 4(-16)(s_{0} - h)}}{2(-16)}$
Solving the above, we get
t $=$ $\frac{ v_{0}±\sqrt {(v_{0})^{2}+ 64(s_{0}-h)}}{32}$
