2. Given that $G$ is the centroid of $\triangle A B C$, $G A=5, G B$ $=12, G C=13$. Then the height from vertex $C$ to side $A B$ of $\triangle A B C$ is $\qquad$
$2.13 \frac{11}{13}$.
Extend $B G$ to intersect $A C$ at point $D$, then $B D$ is a median,
$$
\begin{array}{l}
G D=\frac{1}{2} B G \\
=6 .
\end{array}
$$

Extend $G D$ to point $E$ such that $D E=G D=6$, and connect $A E$.
It is easy to see that $\triangle A D E \cong \triangle C D G$. Therefore,
$$
A E=G C=13 \text {. }
$$

Since $G A^{2}+G E^{2}=5^{2}+12^{2}=13^{2}=A E^{2}$, then $\angle A G E=90^{\circ}$.
Thus, $S_{\triangle A B D}=\frac{1}{2} B D \cdot G A=\frac{1}{2} \times 18 \times 5=45$.
Since $B D$ is the median of $\triangle A B C$, we have
$S_{\triangle A B C}=2 S_{\triangle A B D}=2 \times 45=90$.
It is easy to see that $A B=\sqrt{G A^{2}+G B^{2}}=\sqrt{5^{2}+12^{2}}=13$.
Let the height from $A B$ be $h$, then $S_{\triangle A B C}=\frac{1}{2} A B \cdot h$, i.e., $90=\frac{1}{2} \times 13 h$. Therefore, $h=13 \frac{11}{13}$.