V=IR, so 1A through a 0.5 Ohm wire = 0.5V drop.
Presumably you have one wire for power and one for ground (right?). In that case you'll drop 0.5V in each wire, giving a total drop of 1V.
Therefore, you get (5 - 1 = ) 4V at the load.
The load resistance does enter the calculation too, though. In order for this calculation to be valid, your load would need to have a resistance of exactly 4/1 = 4 Ohms (from R=V/I). If the load resistance is any higher than this, you'll get less than 1A from the supply, and a voltage higher than 4V at the load.
The calculation you need is: V(load) = V(supply) - V(drop), where V(drop) = I * R(wires), and I = V(supply)/[R(wires)+R(load)]
Rearrange & solve as necessary!