Tim Nikias v2.0 wrote:
> 
> Take, for example, the distance an object has dropped in free fall, the
> formula is
> h = g*t^2 / 2
> There's nothing like air stopping the falling in this formula, it's entirely
> simplified, but still relevant. It just assumes that there's no friction. I
> was asking for the same: in a no friction-system, what's the rate of
> slowdown for an object moving uphill?

As always the key to solving a problem is to understand it.  Your formula:

h = g*t^2 / 2

is in the 1-dimensional case you describe the result of the following ideas:

- the position of a point mass at the time T is:

          T
x = x0 + S  v dt
         t=0

(The 'S' being an integral)

- the velocity of the point mass with an acceleration a is:

          T
v = v0 + S  a dt
         t=0

(x0 and v0 are the starting values for position and velocity at T=0)

Now just integrate twice and you get your formula (assuming starting 
conditions v0=0 and x0=0, h=x, g=a).  For the 2D/3D case it is 
essentially the same, just with vectors for position, velocity and 
acceleration of course.  Note you can have arbitrary starting conditions 
- the formulas will work the same way for 'uphill' and 'downhill'.

But don't forget that a is *not* the gravity but the sum of all 
accelerations - or accoring to Newton: F=m*a - the result of all forces 
influencing the particle.  In case of surface contact (slipping) the 
contact force (which has nothing to do with friction) has to be taken 
into account (and it will strongly vary, a=a(t) in the above integral).

Christoph

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