Geometry

We can use Heron's formula to first find the area of the triangle, $\sqrt{s(s-a)(s-b)(s-c)}$. Since the semiperimeter is $17$, we have 

$\sqrt{17(17-15)(17-9)(17-10)} = \sqrt{1904} = 4\sqrt{119}$

as the area of triangle $ABC$. The shortest altitude of a triangle has the longest side as its base, so we have 

$\frac{15h}{2} = 4\sqrt{119}$,

where $h$ is the altitude. Solving for $h$ yields $h = \boxed{\frac{8\sqrt{119}}{15}}$.

^{(I'm not sure if there's a way to solve this without using Heron's, but if there is, please let me know! Thanks.)}