Answer
$\approx 50.6\ \ \mathrm{f}\mathrm{t}/\mathrm{s}$
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:$\displaystyle \quad\frac{v_{0}^{2}}{32}=40$
$v_{0}^{2}=40\cdot 64$
$v_{0}=\sqrt{2560}\approx 50.6\ \ \mathrm{f}\mathrm{t}/\mathrm{s}$
