$$
\begin{aligned}
&\therefore \frac{\mathrm{AB}}{\mathrm{DC}}=\frac{\mathrm{BE}}{\mathrm{CE}} \Rightarrow \frac{\frac{16}{3}}{2}=\frac{\mathrm{x}+\mathrm{y}}{\mathrm{y}} \\
&\Rightarrow 16 \mathrm{y}=6 \mathrm{x}+6 \mathrm{y} \\
&\Rightarrow \mathrm{y}=\frac{3}{5} \mathrm{x}
\end{aligned}
$$
On differentiating both sides w.r.t. t, we get
$$
\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{3}{5} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{3}{5} \cdot\left(-1 \frac{2}{3}\right)
$$
[since, man is moving towards the light post] $=-1 \mathrm{~m} / \mathrm{s}$
Let $z=x+y$
Now, differentiating both sides w.r.t. t, we get
$$
\frac{\mathrm{dz}}{\mathrm{dt}}=\frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{dy}}{\mathrm{dt}}=-\left(\frac{5}{3}+1\right)=-2 \frac{2}{3} \mathrm{~m} / \mathrm{s}
$$
Hence, the tip of shadow is moving at the rate of $2 \frac{2}{3} \mathrm{~m} / \mathrm{s}$ towards the light source and length of the shadow is decreasing at the rate of $1 \mathrm{~m} / \mathrm{s}$.