Rate of movement of shadow tip

Draw the lamppost of height 15 ft at the origin, with the person of height 6 ft standing at distance $x$ from its base, casting a shadow of length $s$. By similar triangles, $\frac{15}{x+s} = \frac{6}{s}$, since both triangles share the same angle from the light source. Cross-multiplying gives $15s = 6x + 6s$, so $9s = 6x$, which simplifies to $s = \frac{2}{3}x$.

Differentiating $s = \frac{2}{3}x$ with respect to $t$ gives $\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt} = \frac{2}{3}(5) = \frac{10}{3}$ ft/s. The shadow tip's position is $x+s$, so its rate of motion is $\frac{dx}{dt}+\frac{ds}{dt} = 5 + \frac{10}{3} = \frac{25}{3}$ ft/s. Thus the tip of the shadow moves away from the lamppost at $\frac{25}{3} \approx 8.33$ ft/s.