Identify the vertices and asymptotes of the hyperbola. $\frac{x^2}{25}-\frac{y^2}{81}=1$

A) vertices: $(-5,0),(5,0)$ asymptotes: $y=-\frac{5}{9}x,y=\frac{5}{9}x$
B) vertices: $(-5,0),(5,0)$ asymptotes: $y=-\frac{9}{5}x,y=\frac{9}{5}x$
C) vertices: $(0,-9),(0,9)$ asymptotes: $y=-\frac{5}{9}x,y=\frac{5}{9}x$
D) vertices: $(-9,0),(9,0)$ asymptotes: $y=-\frac{9}{5}x,y=\frac{9}{5}x$
E) vertices: $(-9,0),(9,0)$ asymptotes: $y=-\frac{5}{9}x,y=\frac{5}{9}x$

The standard form of the hyperbola is $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$, where $(h,k)$ is the center, $a$ is the distance from the center to the vertices along the $x$-axis, and $b$ is the distance from the center to the vertices along the $y$-axis.

Comparing with the given equation, we have $h=k=0$, $a=5$, and $b=9$. Therefore, the vertices are $(\pm5,0)$ and the asymptotes have slopes $\pm\frac{b}{a}=\pm\frac{9}{5}$. Using the point-slope form, the asymptotes pass through the center $(0,0)$ and have equations $y=\pm\frac{9}{5}x$.

Therefore, the correct choice is $\boxed{\textbf{(B)}}$.