A man 2 metres tall walks away

Let AB be the lamp post. Suppose 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,

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

$\Rightarrow \frac{5}{2}=\frac{x+y}{y}$

$\Rightarrow \frac{x}{y}=\frac{5}{2}-1$

$\Rightarrow \frac{x}{y}=\frac{3}{2}$

$\Rightarrow y=\frac{2}{3} x$

$\Rightarrow \frac{d y}{d t}=\frac{2}{3}\left(\frac{d x}{d t}\right)$

$\Rightarrow \frac{d y}{d t}=\frac{2}{3} \times 4.8$

$\Rightarrow \frac{d y}{d t}=3.2 \mathrm{~km} / \mathrm{hr}$