Simple orbital collision problem -- need help!

[dmrogers99]

I'm probably among the least qualified people to answer your question, but since nobody else has, and I think it's interesting, figured I'd chime in.

Why would being a parabola make horizontal speed at apogee negligible?

The velocity vector with respect to the "vertical" component (distance from earth's center), would have zero magnitude, just before it changed sign and began descent. The horizontal velocity vector would remain the same, otherwise it's never going to go very far, in which case, why expend all that energy?

Presumably it didn't just launch straight up and return the same way. At some point it "tipped" along the azimuth of its intended direction of flight and went really fast above the atmosphere where drag isn't as much of a factor.

I'm assuming this isn't a "constant boost" magical rocket, and at some point it's in ballistic flight until such time as it must begin a powered descent to bleed energy and make a controlled landing, presumably to the delight of all the passengers.

Any maneuvers to "flip" the spacecraft are likely to make the flight path not be a "perfect" parabola, but I think it's still largely parabolic.

Altitude is irrelevant to whether or not the flight is "suborbital," velocity is what matters, so 110KM doesn't make a big difference either way. Most of your V is along the flight path, i.e. "horizontal."

For that, you'll have to use the rocket equation applied to your particular spacecraft. Yeah, you're going to get some acceleration due to gravity, but I think your velocity along the direction of flight is the largest component in terms of added energy in the collision.

But I'm just guessing here. I'm no rocket scientist.