I think by definition free falling means there are no other forces (beside gravity) acting on the object.
The horizontal velocity is independent of the acceleration under gravity, so it will keep moving in a stright line parallel to the x axis with nothing stopping it with constant velocoty, so you should take the horizontal velocity as a constant velocity.
Taking the horizontal and vertical velocity together, it's moving with constant velocity along the x axis and accelerating in the y direction so it's forming a parabolic curve.
Let's work an example to explain.
A car drives off a cliff (remember as physicists we don't ask why the car is going off the cliff we want to know its path after it leavs the cliff).
It is travelling at 5 m/s when it leaves the cliff (parallel to the x axis). Clearly its wheels no longer have any grip on the road so it can't accelerate or decelerate in the x axis.
Now let's say this cliff is 500m high. The first thing we want to know is how long does it take to hit the ground (we're only worrying about the y axis and ignoring the 5 m/s along the x axis right now).
Let's make our calculations easy and say acceleration due to gravity is -10m/s^2. So the distance is -500m (that is to say the car travels 500 m straight down) and acceleration is -10m/s^2 (10 m/s^2 straight down).
s = ut + (1/2)at^2, u is initial velocity (down) and t is time. Well when it drove off the cliff it was travelling downwards at 0 m/s so we can ignore the ut bit.
-500m = (1/2)(-10m/s^2) t^2
-1000m / (-10 m/s^2)= t^2
100s^2 = t^2
t = 10s
So it takes 10 seconds to hit the ground from the 500 metre high cliff.
So the question then is how far along the x axis has it gone in 10seconds? Well it left the cliff at 5m/s (along the x axis) so in ten seconds at a constant velocity of 5 m/s it has travelled 5m/s * 10s = 50m. Therefore the rescuers and authorities should look for the wreckage 50 metres from the base of the cliff.
Does this help?