CONTRARIANTHINKER

WTC7

Delta	Video Time	drop time	Measured drop	Free fall in vacuum drop = 16 * t squared
-	10.50
-	11.00	0	0	0
10	11.50		10
20	12.00	1	30	16
30	12.50		60
35	13.00	2	95	64
45	13.50		140
50	14.00	3	190	144
60	14.50		250
50	15.00	4	300	256
60	15.50		360
60	16.00	5	420	400
>=75	16.50	5.5	>=495

Video frames and measurement

Note 11.50 secs the left side is starting to collapse, right side not

Video used

http://www.911research.com/wtc/evidence/videos/docs/wtc_7_cbs.mpg

Equations

$y(t)=-\frac{1}{2}gt^2+v_{y0}t+y_0$

$v_{y0}\,$ is the initial velocity (ft/s).

$v_{y}(t)\,$ is the velocity with respect to time (ft/s).

$y_0\,$ is the initial altitude (m).

$y(t)\,$ is the altitude with respect to time (ft).

$t\,$ is time elapsed (s).

$g\,$ is the acceleration due to gravity (32 ft/s² near the surface of the earth).

http://en.wikipedia.org/wiki/Free-fall

Note my measurement is of DROP, not absolute HEIGHT, and vy0 = 0 (stationary)

I used feet instead of metres. g = 32 feet per second per second

Therefore the equation for gravity fall in a vacuum simplifies to

y(t) (altitude or 'drop') = -1/2 g t²

g = 32

t = 0, 1, 2, 3, 4, 5

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