A man goes $16 \space m$ due west and then $16 \space m$ due north. How far is he from the starting point?

$22.63 \space m$
1. Let us consider a man starting from $A$ and goes from $A$ to $B$ and then from $B$ to $C$, as shown in the figure.
Then, \begin{aligned} AB = 16 \space m , BC = 16 \space m \space and \space \angle ABC = 90^\circ \end{aligned}
2. From right $\Delta ABC$, we have \begin{aligned} AC^2 =&\space AB^2 + BC^2 \\ =&\space [ (16)^2 + (16)^2 ] \space m^2 \\ =&\space ( 256 + 256 ) \space m^2 \\ =&\space 512 \space m^2 \\ \therefore \space AC = &\space \sqrt{ 512 } \space m^2 = 22.63 \space m \end{aligned} Hence, the man is $22.63 \space m$ away from the starting position.