Answer
Prof. Strong throws about $1.41$ times as fast as Prof. Weak.
Work Step by Step
By Exercise $70$, the highest point is $\displaystyle \frac{v_{0}^{2}}{64}$ feet above the starting point.
Let
$v_{0}$ be the initial speed of Prof. Strong
$v_{1}$ be the initial speed of Prof. Weak.
We are given$\quad $
$\displaystyle \frac{v_{0}^{2}}{64}=2\cdot\frac{v_{1}^{2}}{64}$
$v_{0}^{2}=2v_{1}^{2}$
$v_{0}=v_{1}\sqrt{2}\approx 1.41\cdot v_{1}$
Thus: Prof. Strong throws about $1.41$ times as fast as Prof. Weak.
