If a beam of light was shone horizontally, and simultaneously a stone was dropped from the same height, would they both hit the ground a the same time?
Of course on Earth they would not, but let's imagine a land mass large enough for the light not to fire off into space and away from the pull of the land's gravity.
Answer
Given that the stone is in free fall - that is, in a frame following a geodesic - it would have to observe the light emitted in its frame as behaving normally. This means that the stone would not observe the light bend away from a path radially outwards from it, which is a complex way of saying that it must hit the ground at the same time as the stone.
In the comments above, Floris mentioned how the deflection of light predicted by GR is twice what one expects from Newtonian calculations. This is true, however it is the measured deflection from the frame of a non-local observer that it refers to. So an observer standing on the ground would notice the discrepancy in the deflection of light, but the stone would not notice anything, since it is a local observer. If the stone were to see the photon hit the ground at a different time from it, that would violate the weak equivalence principle.
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