A street lamp on a seafront promenade in Lisbon is 6 m above the ground…

Step 1. The link. The lamp, the top of her head and the tip of the shadow lie on one line. Similar triangles give $\dfrac{s}{6} = \dfrac{s - x}{1.5}$. So $1.5s = 6s - 6x$, which gives $s = \dfrac{4x}{3}$.

Step 2. The tip of the shadow. $\dfrac{ds}{dt} = \dfrac{4}{3}\,\dfrac{dx}{dt} = \dfrac{4}{3} \times 0.9 = 1.2\ \text{m/s}$.

Step 3. The length of the shadow. The length is $s - x = \dfrac{x}{3}$, so its rate is $\dfrac{1}{3} \times 0.9 = 0.3\ \text{m/s}$.

Step 4. The distance at $x = 8\ \text{m}$. $s = \dfrac{4 \times 8}{3} = 10.67\ \text{m}$.

Why the speeds do not depend on $x$. The link $s = \dfrac{4}{3}x$ is a straight-line relation, so its derivative is the constant $\dfrac{4}{3}$.

If the lamp were higher. For a lamp of height $H$, the link is $s = \dfrac{H}{H - 1.5}\,x$. The factor $\dfrac{H}{H - 1.5}$ gets closer to $1$ as $H$ grows, so the tip would move more slowly, and almost as slowly as the woman.