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Originally posted by nQuire
No no no.
dy=1/2at^2 will yield one dimensional distance travelled under constant acceleration in time, Valhall's suggestion only adds initial velocity. Horizontally, you're looking at a [rapid initial acceleration which we'll best not consider here and subsequent] constant velocity, yielding a horizontal displacement of dx=vt
First equation for dy=400m yields t=9s which you then enter into the second yielding dx=v times 9s. You see we're missing a variable here. Using for example dx=200m (how distant was the Wintergarden?), v would have been 22m/s.
[Disclaimer: These rules apply only in a vacuum and under standard earth acceleration]
In case anyone's confused:
[edit on 14-9-2006 by nQuire]
Can anyone tell me how I would calculate how far the steel can reach without any force other than gravity when it falls from a height of lets say 400 meters?
Originally posted by wecomeinpeace
For horizontal launch, the formula to work out distance achieved is:
d = v*SQRT(2h/g)
Where,
v is lateral velocity at launch
h is height from the ground
g is acceleration due to gravity
As Val said, gravity won't affect the horizontal displacement. I assume you hope to prove that the collapse of the twin towers wouldn't be sufficient to hurl the debris as far as was observed if gravity was the only force acting in the system. In order to do that, you'd have to perform an individual case study on a piece of debris. You'd need to know the exact mechanism by which the piece was ejected horizontally, and then show how that mechanism could not provide sufficient impulse to accelerate the debris to the required launch velocity. As you can see, nowhere near as simple as you perhaps first assumed.