A man 2 metres high walks at a uniform speed of 6 km/h

Let AB be the lamp post. Let at any time t, the man CD be at a distance of x km from the lamp post and y m be the length of his shadow CE.

Since triangles $A B E$ and $C D E$ are similar,

$\frac{A B}{C D}=\frac{A E}{C E}$

$\Rightarrow \frac{6}{2}=\frac{x+y}{y}$

$\Rightarrow \frac{x}{y}=\frac{6}{2}-1=2$

$\Rightarrow \frac{d y}{d t}=\frac{1}{2} \frac{d x}{d t}$

$\Rightarrow \frac{d y}{d t}=\frac{1}{2}(6)$

$\Rightarrow \frac{d y}{d t}=3 \mathrm{~km} / \mathrm{hr}$