Answer
a) $\sqrt{gL}$
b) $5.86\;\rm mph$
Work Step by Step
a) As the author told us, your body will take a circular path (take its center of mass) around the forward foot. This means that when the center of mass of your body is just above your forward foot the radial acceleration will be downward, as seen in the figure below.
Now we need to use Newton's second law to find the maximum speed of your body.
$$\sum F_r=mg-F_n=ma_r=m\dfrac{v^2}{r}$$
Thus,
$$mg-F_n=m\dfrac{v^2}{r}$$
As the author told us, the normal force decreases while this rotation process of your center of mass as your speed became faster and faster.
So, we can find the maximum speed at the minimum normal force which is zero. So, let's say that $v_{max}$ occur when $F_n=0\;\rm N$
$$mg-0=m\dfrac{v_{max}^2}{r}$$
where $r$ here is the length of your leg.
$$ \color{red}{\bf\not}mg = \color{red}{\bf\not}m\dfrac{v_{max}^2}{L_{leg}}$$
$$\boxed{v_{max}=\sqrt{gL_{leg}}}$$
It is obvious that the maximum speed depends on the length of the leg which means that the taller person runs faster than the shorter one which is a reasonable result.
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b)
Plugging the known;
$$v_{max}=\sqrt{9.8\times 0.7}=\color{red}{\bf 2.62}\;\rm m/s$$
$$v_{max}=\color{red}{\bf 2.62}\;\rm m/s= \color{red}{\bf 5.86}\;\rm mph$$
Since the normal walking speed is 3 mph, so this answer of about 6 mph is reasonable when running.
