5. In $\triangle A B C$, $A B=A C=5, B C=6$, and its orthocenter $H$ satisfies $\overrightarrow{A H}=m \overrightarrow{A B}+n \overrightarrow{B C}$. Then $m+n=$ $\qquad$.
5. $\frac{21}{32}$.

Let the altitude from $A$ to $BC$ be $AD$, and draw $HE \parallel BC$ intersecting $AB$ at point $E$. Then $AD=4$,
$$
HD=BD \tan \angle DBH=BD \cot \angle ACB=3 \times \frac{3}{4}=\frac{9}{4}.
$$

Thus, $m=\frac{AE}{AB}=\frac{AH}{AD}=\frac{7}{16}, n=\frac{BH}{BC}=\frac{BH}{2BD}=\frac{7}{32}$, so $m+n=\frac{21}{32}$.