In right triangle $JKL$ with $\angle JLK = 90^\circ$, we have $KL = 8$ and $\tan K = \frac{15}{8}$.  Find $JL$.
[asy]
pair J,L,K,M;
J = (0,8);
K = (15,0);
L = (15,8);
M = (15,4);
draw(J--K--L--J);
draw(rightanglemark(J,L,K,23));
label("$K$",K,SE);
label("$J$",J,NW);
label("$L$",L,NE);
label("$8$",M,E);
[/asy]

We know that $\tan K = \frac{JL}{KL} = \frac{JL}{8}$. Then $\frac{JL}{8}=\frac{15}{8}$, so $JL = \boxed{15}$.