In triangle $ABC$, we have that $AB=AC$, $BC=16$, and that the area of $\triangle ABC$ is $120$. Compute the length of $AB$.
1. Given that \( \triangle ABC \) is isosceles with \( AB = AC \) and \( BC = 16 \), we can drop a perpendicular from \( A \) to \( BC \) at point \( D \), which bisects \( BC \). Therefore, \( BD = DC = \frac{BC}{2} = \frac{16}{2} = 8 \).

2. The area of \( \triangle ABC \) is given as \( 120 \). The area of a triangle can also be expressed as:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   Here, the base \( BC = 16 \) and the height is the altitude from \( A \) to \( BC \), which we denote as \( h \). Therefore:
   \[
   120 = \frac{1}{2} \times 16 \times h
   \]
   Solving for \( h \):
   \[
   120 = 8h \implies h = \frac{120}{8} = 15
   \]

3. Now, we need to find the length of \( AB \). In \( \triangle ABD \), we can use the Pythagorean theorem. Since \( AD \) is the altitude, \( AD = h = 15 \) and \( BD = 8 \):
   \[
   AB^2 = AD^2 + BD^2
   \]
   Substituting the known values:
   \[
   AB^2 = 15^2 + 8^2 = 225 + 64 = 289
   \]
   Taking the square root of both sides:
   \[
   AB = \sqrt{289} = 17
   \]

The final answer is \( \boxed{17} \).