Let $\mathrm{AB}$ be the ladder and $\mathrm{OB}$ be the wall. At an instant, let $\mathrm{OA}=\mathrm{x}, \mathrm{OB}=\mathrm{y}$,
$x^2+y^2=25 \quad \ldots(i)$
On differentiating,
$\begin{aligned}
&2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0 \\
&\Rightarrow x \frac{d x}{d t}+y \frac{d y}{d t}=0 \quad \ldots(ii)
\end{aligned}$
When $x=4$, then from (i), we have $16+y^2=15 \Rightarrow y=3$
Now, $\frac{\mathrm{dx}}{\mathrm{dt}}=0.02 \mathrm{~m} / \mathrm{sec}$. Put these values in (ii), $4 \times 0.02+3 \times \frac{\mathrm{dy}}{\mathrm{dt}}=0 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{-0.08}{3}=\frac{-2}{75}$
Hence, the height of the ladder on the wall is decreasing at the rate of $\frac{2}{75} \mathrm{~m} / \mathrm{sec} .=\frac{8}{3} \mathrm{~cm} / \mathrm{sec}$.