Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola: $y^{2}-16x^{2}=16$.

(N/A) Dividing the equation by $16$ on both sides,we get $\frac{y^{2}}{16}-\frac{x^{2}}{1}=1$.
Comparing this with the standard equation $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$,we find $a^{2}=16$ and $b^{2}=1$,so $a=4$ and $b=1$.
Calculating $c$: $c = \sqrt{a^{2}+b^{2}} = \sqrt{16+1} = \sqrt{17}$.
The coordinates of the foci are $(0, \pm c) = (0, \pm \sqrt{17})$.
The coordinates of the vertices are $(0, \pm a) = (0, \pm 4)$.
The eccentricity $e = \frac{c}{a} = \frac{\sqrt{17}}{4}$.
The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2(1)}{4} = \frac{1}{2}$.