Answer
Prof. Weak throws about $1.73$ times as fast as Prof. Strong.
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$\displaystyle \quad \frac{v_{1}^{2}}{64}=3\cdot\frac{v_{0}^{2}}{64}$
$v_{1}^{2}=3v_{0}^{2}$
$v_{1}=v_{0}\sqrt{3}\approx 1.73\cdot v_{0}$
Thus: Prof. Weak throws about $1.73$ times as fast as Prof. Strong.
