Solution:

If $A$, $B$, and $C$ are the tops of the heads of three successive people and $D$, $E$, and $F$ are their respective feet, let $P$ be the foot of the perpendicular from $A$ to $BE$ and let $Q$ be the foot of the perpendicular from $B$ to $CF$. Then, by equal angles, $\triangle ABP \sim \triangle BCQ$, so
$$\frac{CF}{BE} = \frac{CF}{BQ} = \frac{CQ}{BQ} + 1 = \frac{BP}{AP} + 1 = \frac{BE}{AP} = \frac{BE}{AD}$$
Therefore the heights of successive people are in geometric progression. Hence, the heights of all five people are in geometric progression, so the middle height is $\sqrt{3 \cdot 7} = \sqrt{21}$ feet.