Answer
$\approx 35.8\ \ \mathrm{f}\mathrm{t}/\mathrm{s}$
Work Step by Step
By Exercise $70$, the highest point is $\displaystyle \frac{v_{0}^{2}}{32}$ feet above the starting point.
We are given:$\displaystyle \quad\frac{v_{0}^{2}}{32}=20$
$v_{0}^{2}=20\cdot 64$
$v_{0}=\sqrt{1280}\approx 35.8\ \ \mathrm{f}\mathrm{t}/\mathrm{s}$
