Don't know what combining horizontal and vertical velocities have to do with this, but...
Assuming distance in the x direction, then: d=(cos Vx)(t). Clearly from this equation there is a direct coorelation between the magnitude of vector Vx and d, therefore the higher values for cos should be the answer. If we think about this conceptually, throwing a pitch side arm almost parallel to the ground (for max value of cos=1 at angle of 0 degrees) would yield a further x distance than throwing it up in the air at a lesser angle--assuming equal velocities, of course.
because wiith the 45 degree angle we are going to have an optimal blend between horizontal and vertical velocities
Vf=Vo+a*tA 45 degree angle is clearly the magic angle that maximizes travel time and distance.
I agree with IntelInside... a 45 degree angle is the optimal blend. Here's an example (slightly modified) taken from EK 1001
A projectile is launched at an angle of 30 degrees to the horizontal. It's initial velocity is 20m/s. How far does it travel?
To solve this you have to first find total flight time. Using v=vo+at where v=0m/s at the top of the flight path; vo=vo(sin30); and a=-10m/s
We see that t=1 sec
That's only for half the trip though so total flight time is t=2 sec
Now to find total distance traveled we can use the equation d=vot+1/2at^2
Where a=0 because there's no acceleration in the x-direction; t=2 sec; and vo=vo(cos30)
We see that d=36m
Now, do the same problem with a 45 degree angle.
We see that t=3.4 sec and d=57.8m
Well, what about a 60 degree angle?
We see that t=3.6 sec and d=36m
A 45 degree angle is clearly the magic angle that maximizes travel time and distance.