jmc

We use the distance formula: $$\begin{array}{rl}\sqrt{(-15-0{)}^2+(8-0{)}^2}& =\sqrt{225+64}\\ & =\sqrt{289}=\boxed{17}.\end{array}$$ - OR -

Note that the origin, the point $(-15,8)$, and the point $(-15,0)$ form a right triangle with legs of length $8$ and $15.$ This is a Pythagorean triple, so the length of the hypotenuse is $\boxed{17}$.