A man of height h = 1.8 m walks at a constant speed v = 0.75…

A man of height $h = 1.8\ \mathrm{m}$ walks at a constant speed $v = 0.75\ \mathrm{m/s}$ away from a streetlight mounted at a height $H = 4\ \mathrm{m}$ above the ground. What is the speed of the tip of his shadow along the ground?

Step 1: Set up the geometry

Let $x$ be the man's distance from the foot of the lamp post and $s$ the distance of the shadow's tip from the same point. The ray from the lamp that just grazes the man's head hits the ground at the tip of the shadow.

Step 2: Similar triangles

The large triangle (lamp, foot of the post, tip) and the small triangle (head, feet, tip) are similar, so height divided by base is the same in both: $\frac{H}{s} = \frac{h}{s - x}$.

Cross-multiply and solve for $s$: $H(s - x) = h s \;\Longrightarrow\; s\,(H - h) = H x$.

\[ s = \frac{H}{H - h}\,x = \frac{4}{2.2}\,x = \frac{20}{11}\,x. \]

Step 3: From positions to speeds

The tip is always $\tfrac{20}{11}$ times as far from the post as the man. So whenever the man moves by $\Delta x$, the tip moves by $\tfrac{20}{11}\Delta x$ in the same time. Dividing by $\Delta t$ (or taking the derivative with respect to time):

\[ v_s = \frac{\Delta s}{\Delta t} = \frac{H}{H - h}\cdot\frac{\Delta x}{\Delta t} = \frac{H}{H - h}\,v. \]

Step 4: Substitute the values

\[ v_s = \frac{20}{11} \cdot 0.75\ \mathrm{m/s} = \frac{15}{11}\ \mathrm{m/s}. \]

\[v_s = \frac{15}{11}\ \mathrm{m/s} \approx 1.36\ \mathrm{m/s}\]

Check with two positions

$ x = 3\ \mathrm{m}: \quad s = \tfrac{20}{11} \cdot 3 \approx 5.45\ \mathrm{m} $ one second later, $ x = 3.75\ \mathrm{m}: \quad s = \tfrac{20}{11} \cdot 3.75 \approx 6.82\ \mathrm{m} $ the tip moved $ 6.82 - 5.45 \approx 1.36\ \mathrm{m} \text{ in } 1\ \mathrm{s}\ \checkmark $