Rate of change of shadow length

Draw the pole of height 15, the person of height 6 standing a distance $x$ from the pole, and the shadow of length $s$ extending beyond the person. By similar triangles, $\frac{15}{x+s} = \frac{6}{s}$, since both triangles share the same angle at the tip of the shadow. Cross-multiplying gives $15s = 6x + 6s$, so $9s = 6x$, which simplifies to $s = \frac{2}{3}x$.

Differentiate both sides with respect to $t$: $\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt}$. Since the person walks away from the pole at $\frac{dx}{dt} = 5$ ft/s, substitute to get $\frac{ds}{dt} = \frac{2}{3}(5) = \frac{10}{3}$ ft/s. Note the answer does not depend on $x = 10$, since $s$ is proportional to $x$.