Lunar Hammer, Part 1 - Escaping the Moon

I had a whole different post written up for this, only to find that a statement I came across recently, and believed to be balony, does seem to be true:

"Escape velocity is independent of the angle of projection".

Basically, it doesn't matter which direction you shoot something, if it equals or exceeds Escape Velocity, then it will keep on going out of the local gravity well. This condition only applies if there is no atmosphere to complicate things, however. But, happily, I am concerned with the Moon. What I do find bizarre is how this bears little resemblance to Newton's mountaintop cannon thought experiment (cannonballs being fired tangentially to the Earth's surface). How can any old angle lead to an escape trajectory? But an equation, found on Wiki, clearly implies this (about halfway down, just above 'trajectory'. Frankly, to my brain still cached with school days projectile maths, it seems somewhat unreal. I do not have much gift for math, but there it was: as 'v' > EV, then 'h' > infinity. It shows, to be more precise, an asymptotic relationship between height and velocity.

Not knowing all this doesn't actually have any effect on my Lunar Hammer system though. As a result of (now erroneous) feedback from the BIS, I arrived at a solution that will work regardless for angles up to 45 degrees from the vertical. But now, I find that Hammer is free to do something amazing. More coming on that in near future posts.