Tim Nikias wrote:
> 
> Well, I don't understand, but I also think that you've misunderstood my
> (or, which would be more precise, I didn't ask too well ;) ) problem.
> 
> This is the environment:
> -Gravity, which pulls downward
> -no wind
> -no air friction
> 
> This is the particle:
> -Speed
> -Position
> -Target
> 

So, When you say 'flying', it is just 'ballistic' ? (initial speed then
get more gravity, nothing else) 
Or is it rather 'propulsion' (there is some continuous thrust as well as
gravity).


> Everything takes place in 3D, but, since the path could be aligned to some plane
> or another, we can stay put with 2D.


> I can calculate the relation between position and target as range, as absolut
> distance,
> and as height.

Ok, in 2d, ballistic, easier, first:

   y = a.t.t + b.t + c
   x = d.t

 (t = clock !)

you will have to resolve the system for the initial speed vector
 ( dx & dy at t=0 ), using the target and initial position.
 There is an infinity of solution without the speed, because 
 the curve is a parabol and you have only two poind.
 speed fix the tangent and the length of speed should be
 adjusted according to the gravity intensity.

Beware: the model assument a plane world, not a spherical planet.
 (experiment during WWI & WWII show that earth spherical nature has
 an impact on canonball destination).

Propultion:
 Question: can the direction of the thrust change during the 'fly' ?
   (including can the thrust be turn on/off instantly)
 if yes and the thrust is stronger than the gravity, any curve are possible.
 if yes and the thrust is not stronger, it can be very complex, unless
  you simply use the thrust to either:
     - first gain horizontal velocity, then only to counter-act gravity
          (go back to the ballistic model, with a reduced gravity for the
           second step. For the first step, same ballistic model, but
           x = d.t + e*t*t/2 )
     - only try to counter-act part of the gravity
     - only gain constant horizontal acceleration
           (ballistic model, with x = d.t + e*t*t/2 )
 if no: the vertical component of the trust just modify the local gravity effect
         (reduce g), while the horizontal part keeps adding acceleration
             (see above)

> 
> Question is: Which angle does the particle's thrust (speed) has to be in
> relation to the ground plane, in order to hit the target?

For the ballistic model:
  for a given gravity and target, if the angle is free, there is potentially
 an infinite number of answer (when there is an answer), unless the amount of
 speed is somehow limited.

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