I think you have an inaccurate understanding of how projectile motion works.
First, there is no particular angle that you need to fire to "escape the horizon". The horizon is not a well-defined distance (it depends on how high how above sea-level you are), and in any case you never need to fire at a steeper angle than 45 degrees.
Second, momentum is not "used up" to elevate a projectile. If you ignore air resistance, a cannon shell will hit the ground with exactly the same speed as it was launched (no matter what angle it was launched at). For example, if you fire it straight up, then the kinetic energy of the projectile when it lands (again, disregarding air resistance) is exactly the height it reached times the force of gravity. And that is exactly the speed it had when it was initially fired (which is how it could reach that height).
> First, there is no particular angle that you need to fire to "escape the horizon".
You cannot make up fictional quotes then chastise me for my lack of understanding based on those quotes. I'm not even going to respond to that.
> Second, momentum is not "used up" to elevate a projectile.
That is where you're mistaken. The higher the angle that the projectile is launched at the more energy that gets transferred into elevation.
So when firing straight forward you transfer almost no energy from forward momentum into height, which is why when you stand straight, stick out your arm, and fire at a target down range you have to adjust for gravity.
As you point the "gun" further and further upwards the "bullet" will travel higher and higher but start to hit the ground at no more than its terminal velocity. This is because energy is going towards height from forward speed until forward speed is near zero.
> If you ignore air resistance, a cannon shell will hit the ground with exactly the same speed as it was launched (no matter what angle it was launched at).
Then where does the energy come from in order to gain their altitude? If we take what you just said literally, then a mortar is an impossible kind of weapon, since you have no way of making the projectile travel upwards, only forwards.
The problem with that is that you just contradicted your own post. Earlier you said that 45 degrees was the optimal angle for distance, correct? So you acknowledge that at different degree elevations projectiles travel different distances, would this also be correct? Therefore if we combine these two points together: that different degrees cause projectiles to travel different distances, and that that travel requires energy, you must acknowledge that that energy has to come from the projectile's forward momentum.
If you acknowledge that forward momentum is converted into height at higher degree elevations (as you did earlier, see 45 degree point) you must also acknowledge that energy is being eaten for this additional height.
> For example, if you fire it straight up, then the kinetic energy of the projectile when it lands (again, disregarding air resistance) is exactly the height it reached times the force of gravity.
That is incorrect. If you fire a projectile straight up it doesn't come down "times the force of gravity." In fact things cannot exceed their terminal velocity which is 1x gravities unless they're being artificially accelerated (e.g. firing a rocket straight down can go beyond 1x gravities, but a falling brick cannot exceed 1x gravities).
The image you posted in the sister comment made me see what's going on. You are arguing according to the Aristotelian theory of projectile motion, while the other commenters assume the Galilean theory. See e.g. this link: http://www.met.reading.ac.uk/pplato2/h-flap/phys2_2.html#sec... .
(In the presence of strong air resistance, the Aristotelian theory is not a bad approximation. If there is air resistance, a falling object will indeed gradually slow down to a terminal velocity, like you write in your last paragraph.)
First, there is no particular angle that you need to fire to "escape the horizon". The horizon is not a well-defined distance (it depends on how high how above sea-level you are), and in any case you never need to fire at a steeper angle than 45 degrees.
Second, momentum is not "used up" to elevate a projectile. If you ignore air resistance, a cannon shell will hit the ground with exactly the same speed as it was launched (no matter what angle it was launched at). For example, if you fire it straight up, then the kinetic energy of the projectile when it lands (again, disregarding air resistance) is exactly the height it reached times the force of gravity. And that is exactly the speed it had when it was initially fired (which is how it could reach that height).