When you launch a projectile horizontally, it does not undergo an horizontal acceleration, but it does undergo a vertical acceleration that is the same as when the projectile is simply dropped.
If you are interested in the math, here you go:
For launches at an angle, the initial velocity has x and y components as:
Vx = Vcos(Θ) ; horizontal component
Vy = Vsin(Θ) ; vertical component
The equations below can be derived from the basic equations of motion.
Max height: Irrelevant for horizontal launch
h = Vy²/2g
h = v²sin²(Θ)/2g
Time in flight: same as dropped object
t = √2h/g ; this is a different equation on a trajectory
Horizontal distance:
d = Vcos(Θ)t ; this is a different equation on a trajectory
d = Vcos(0)t ; cos(0) = 1
d = Vt
d = V√2h/g
I cut out a lot of steps, but this should give you an idea.