$cos\angle C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac{5^2+5^2-6^2}{2\cdot5\cdot5}=\frac7{25}$

$sin\angle C=\sqrt{1-cos^2\angle C}=\sqrt{1-\left(\frac7{25}\right)^2}=\frac{24}{25}$

$sin\angle C=\frac{24}{25}=\frac{AH}{AC}$

$AH=AC\cdot sin\angle C=5\cdot\frac{24}{25}=4,8$