Is there any way a meteor can hit at less than escape velocity? Note:  I have figured out one case:  The meteor is splash from an impact on a close-in moon.  I'm looking for any path from deep space.

My understanding is that anything from deep space will inherently pick up the escape velocity in it's fall and thus there is no possibility in the simple case.  I'm not so sure about what could happen after an encounter with one or more moons, though.
Edited.
No if you talk about the escape velocity form Earth. This follows simply from the fact that energy $E$ is conserved. An object that is not gravitationally bound to Earth has $E&gt;0$ and hence $v&gt;v_{\mathrm{esc}}$ when hitting ground.



Yes, if you meant the escape velocity from the Solar system, because the Earth moves with $v_{\rm Earth}=v_{\rm escape}/\sqrt{2}$ relative to (but not towards) the Sun. Here $v_{\rm escape}=\sqrt{2GM_{\odot}/1{\rm AU}}$ is the local escape speed from the Sun, while $v_{\rm Earth}=\sqrt{GM_{\odot}/1{\rm AU}}$ is the speed of the local circular orbit. An object at 1AU form the Sun and bound to the Sun cannot have speed greater than $v_{\rm escape}$.

Now, the impact speed of an object that moves at $v_{\rm escape}$ can be as low $v_{\rm escape}-v_{\rm Earth}=v_{\rm escape}(1-1/\sqrt{2})$ if it hits Earth "from behind", i.e. moving in the same direction as Earth at the time of impact.

Note also that meterors typically move not faster than $v_{\rm escape}$, for they don't come from outer space, but from the Solar system.