Bacterius, correct me if I'm wrong:

Your equations assume:

1) fixed starting point,

2) fixed ending point,

3) fixed gravity and

4) no restrictions as to how much time the object needs to take there.

Using that information, you derive the initial necessary velocity to move from point A to point B.

Assumptions 1) and 2) apply here as well, but from the initial post, as I understand, the gravity is an unknown which must be found, and you're given the total time T which the movement takes (or needs to take).

I don't believe that your equations, as you posted them in the current format, will solve these requirements.

Of course, I could be missing something...

The parametric equation relates the three variables gravity, time from A to B and initial velocity, and can be rearranged to express any one of those variables in terms of the other two. Restrictions can be placed on any variable, including time. The original post gives a required time which can be plugged in, yielding equation relating g and initial velocity v0. With no restrictions, infinitely many (g, v0) solutions exist, and can be found by simply choosing arbitrary g or v0 and solving for the other.

Unless I made a mistake, which is entirely possible.

Two of those are unknown in the problem posted in this thread - the original velocity and the gravity.
Perhaps I'm missing the obvious, but there's no trivial way to expand or rearrange the equation to solve for both gravity and initial velocity -only a way to give you a relation between them (two unknowns, one equation). This would usually mean that there might be 0, or more solutions to the equation, however the one piece of information you do not account in those equations is the maximum height that the object needs to reach - which I think removes the last degree of freedom and narrows down the solution to only 1.

Ah, you are correct, I did not see that information. This will indeed be enough. In this case my equation applies twice, once at maximal height at time t, and once at initial height at double the time (2t), and you have to solve a system of two equations in g and v0 which should give the same (unique) solution you obtain. Thanks.

Easiest way is to equate initial velocity, and solve for g, and use that to solve for initial velocity (like you did) which gives the same solution.

Assuming initial height is the ground, this will for instance give solution v0 = 4h / t which is equivalent to your solution assuming Ha = 0.

At this point your solution probably makes more sense to everyone, though :)