gravity - Is there any way a meteor can hit at less than escape velocity?

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 $E>0$ and hence $v>v_{mathrm{esc}}$ when hitting ground.

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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.

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