You asked a related question in another post, and you erased the question while I was posting an answer. So I post it here. The question was complementary to the one above so I think it's relevant to include the answer: why are non-uniform lattices in rank 1 symmetric spaces of noncompact type not hyperbolic except in the case of the hyperbolic plane? Here is my answer.
I think it's a result of Garland and Raghunathan (Annals 1970 "Fundamental domains..."). They show that given such a lattice, for some point $\omega$ at infinity, the stabilizer of $\omega$ in the lattice acts cocompactly on each horosphere based at $\omega$. This horosphere is modeled on a $(n-1)$-dimensional simply connected nilpotent Lie group, where $n$ is the real dimension of the rank 1 symmetric space of non-compact type. Thus the nonuniform lattice contains a f.g. nilpotent group of Hirsch length $n-1$. This is possible in a hyperbolic group only if $n\le 2$. (Note: if $n\ge 3$, it follows that the lattice contains a free abelian group of rank 2.) More precise results about the structure of these lattices were formalized into the concept of relatively hyperbolic groups, see Gromov, Farb, etc. They indicate that intuitively, these "peripheral" subgroups are the only obstruction to hyperbolicity.