I know I kinda did a bad job of explaining that initially, so fuck words. Gonna try straight math. I split it into parametrics because it's slightly easier to calculate that way.
z=total velocity at the lip, on the x-y plane. Like x/y.
theta= lip angle
vertical v(y-initial)=z*sin(theta)+ pop
horizontal v(x-initial)= z*cos(theta)
So, for the position equations,
y=-4.9*t^2 + v(y-initial)*t +0
x= v(x-initial)*t
So as theta increases, v(x-initial) decreases and v(y-initial) increases. No matter what, given the same horizontal velocity, you'll go further with a greater v(y-initial). If you're popping right, you're pushing straight up, and barely at all against your horizontal speed. This is supported empirically, as well, since popping harder always seems to make you go further. Keep in mind that this explanation is only good for lip angle under 45 degrees. Above that popping doesn't help you.
Actually after thinking about this some more, you're right, but only in this special case, which is pretty unlikely. It only works if you're hauling ass and absorb the shit out of the jump, and if the landing is super steep--which is exactly what you describe. Like it would have to almost look like a super g jump. Basically, when you absorb like that, you're reaching the same point in space at the same time, but you're doing it with a smaller vertical velocity, so you still have further to go before you hit the landing. A skim effect. And making a steeper kicker would counteract that.
Dammit, I need to do my math homework, but this is too interesting!