It's a scientific fact

All bodies free fall at the same constant rate (about
16.1'/sec^2=s/sec^2=g/2); yet SI has no units that make use of this
constant: If they did the kilogram would have units of
[(2w/g)=weight/(g/2)]; or one kilogram equals two newtons/g!

"Don1" <[Only registered users see links. ]> wrote in message
news:1111374979.158770.162860@l41g2000cwc.googlegr oups.com...

Oh I thought it was going to be something interesting.

The rate of free fall is not a constant. The rate of fall depends on
the strength of the gravatational field. objects of differing masses
fall at the same rate in any one field, but at different rates in
different field strengths.

[Only registered users see links. ] wrote:

You're right max! Good going; but this got a lot of peoples' attention:
I should have been more specific, and said something more like:

At any particular time and place on Earth, in an environmentally
controled laborarory, all bodies free fall at the same constant rate
(about 16.1'/sec^2=s/sec^2=g/2); yet SI has no units that make use of
this constant: If they did the kilogram would have units of
[(2w/g)=weight/(g/2)]; or one kilogram equals two newtons/g!

Don

Don1 wrote:

I can't think of an interpretation that makes this correct.
Please define what you mean by the "rate of fall".

PD

"PD" <[Only registered users see links. ]> wrote in message
news:1111420647.924961.263710@f14g2000cwb.googlegr oups.com...

I think he is saying if you are for example at the top of the Leaning Tower
of Pisa and release your balls they will fall at a constant rate to the
ground

SR

Steve Ralph wrote:
Tower
the

That *still* is not precise enough. Here are some choices:
1. The time elapsed for the bodies to fall the same distance is the
same.
2. The final velocites the bodies have at impact will be the same.
3. The rate of change of the velocities (the acceleration) of the
bodies will be the same.
4. The average velocity of the bodies over the distance of the fall are
the same.

What quantity is he referring to here?

PD wrote:
Whereas the acceleration of free fall (g), here on Earth is the average
time rate of change, in a rate of change in position, where
g=(vt-vi)/(2t)=32.174'/sec^2; the rate of free fall is (more simply)
the rate of change in position, where g/2=s/t^2=16.08'/sec^2.

Don

"PD" <[Only registered users see links. ]> wrote in message
news:1111429787.296264.281950@g14g2000cwa.googlegr oups.com...

The rate of velocity, more commonly know as acceleration.

Dwayne

"Steve Ralph" <[Only registered users see links. ].uk> wrote in message
news:423ef651$0$15989$[Only registered users see links. ].net...
Tower
the

I'd see a doctor before it got THAT bad.