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

dB

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