
The Calculators will not be functional here you will have to go to the site

See the method for used for estimating distances http://www.v911t.org/MakingScale.php




The original 7 World Trade Center was a 47story building, designed by Emery Roth & Sons, with a red granite façade. The building was 610 feet (186 m) tall, with a trapezoidshaped footprint that was 330 ft (101 m) long and 140 ft (43 m) wide.[1][2] Tishman Realty & Construction managed construction of the building, which began in 1984.[1] In March 1987, the building opened, to become the seventh structure of the World Trade Center. http://en.wikipedia.org/wiki/7_World_Trade_Center  


So we see that 345 feet is covered in 5.125, and 375 feet is covered in 5.33 seconds. Below are the calculations of distance covered in freefall in 5.125 and 5.33 seconds.


Freefall at 5.125 secs = ft and Freefall at 5.33 secs. = ft so it is safe to say that WTC 7 dropped at or very near freefall velocity. 422 v 345 (81.75%), is 77 feet more distance traveled. 456 v 375 (82.23%) freefall is 81 ft more distance traveled, so we can see based upon the gain of the % comparison that as the fall continues the building is getting closer to freefall velocity, not further away, in other words resistance is decreasing. WTC 7 is somehow accelerating at an increasing rate.
See the method for used for estimating distances http://www.v911t.org/MakingScale.php 

This object falls 100 Meters (328 ft) in 4.4 seconds very nearly the same rate of fall as the examples above, further evidence of falling with nearly no resistance. 
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CD chart courtesy of NASA:
http://exploration.grc.nasa.gov/education/rocket/shaped.html
NASA has neglected to tell us what the frontal area is.
CD for a sphere has a huge range from .07 to .5
CD @ .07 dropped from 610 feet = TV of 290.274 ft/sec = fall time of 6.1599
CD @ .5 dropped from __610 feet = TV of 108.61 ft/sec_ = fall time of 7.3043 secs
CD @ .7 dropped from _610 feet = TV of 91.792 ft/s ___= fall time of 8.077 secs
Apples are not perfect spheres and so without wind tunnel testing one could estimate that .5 is closer to the CD of an apple than .07, or if we dropped an apple from 610 feet we could back our way in to the CD with algebra if we could observe the fall time. My guess is that the CD for an apple is more like .7 depending on the apple. Using a of .7 CD @ .7 dropped from __610 feet = TV of 91.792 ft/sec
Conclusion: Call in the Myth Busters
Freefall Of an Apple to Terminal Velocity, and then fall 610 feet (equal to WTC 7)
Weight of Apple .43 Pounds
Size of Apple is 2.75 x 3 inches = 0.0625 sq ft
Drag Coefficient = .5 (worst case aerodynamic for a sphere) Note that an apple is not a sphere
Drag Coefficient = .07 (best aerodynamic case for a sphere)
Source:
http://exploration.grc.nasa.gov/education/rocket/shaped.html
Terminal Velocity = 290.274 ft per second or 88.4755152 meters per second. Source:
http://exploration.grc.nasa.gov/education/rocket/termvr.html
=========================== Using 0.07 CD and 610 Feet ===============
At time t = s after being dropped, the apple reaches terminal velocity and stops accelerating.
the speed is v_{y} = m/s = ft/s , = terminal velocity
The distance from the starting point will be
y = m= ft here shows you would need 1310 ft to accelerate to terminal velocity, so the apple would fall at freefall for 610 feet. which is seconds and never achieve terminal velocity.
=========================== Using .5 CD and 610 Feet================
After 3.3780 seconds, after travailing a distance of 183.4432 feet the apple reaches terminal velocity, then would travel another 426.5568 ft at108.61 ft/s = 3.9263 + initial time of 3.3780 = 7.3043, total distance of 610 feet
The National Institute of Standards and Technology (NIST) states the height of WTC7 as 610ft = 185.928m
At time t = s after being dropped, the apple hits terminal velocity
the speed is v_{y} = m/s = ft/s , TV is achieved at after
The distance from the starting point will be
y = m= ft In the first 3.3780 seconds the apple would travel 183.4432 feet with 426.5568 feet to go @ 108.61 ft/s = 3.9263 secs + initial time of 3.3780 secs = 7.3043 secs
=========================== Using .7 CD and 610 Feet================
At time t = s after being dropped, the apple hits terminal velocity
the speed is v_{y} = m/s = ft/s , terminal velocity
The distance from the starting point will be
y = m= ft In the first 2.8549 seconds the apple would travel 131.0302 feet with 478.9698 feet to go @ 91.792 ft/s = 5.2179 secs + initial time of 2.8594 secs = 8.077 secs
http://hyperphysics.phyastr.gsu.edu/hbase/traj.html
Alfons v911t
Freefall
In the absence of frictional drag, an object near the surface of the earth will fall with the constant acceleration of gravity g. Position and speed at any time can be calculated from the motion equations.
Illustrated here is the situation where an object is released from rest. It’s position and speed can be predicted for any time after that. Since all the quantities are directed downward, that direction is chosen as the positive direction in this case.
At time t = s after being dropped,
the speed is v_{y} = m/s = ft/s ,
The distance from the starting point will be
y = m= ft
Enter data in any box and click outside the box.
You can play with this and other calculators @ the gsu site to prove the WTC towers fell too fast. I put this here on the v911t for referenece to it, and to have it documented in case.
========== A study of top 15 floors of WTC Tower Fall Time in Atmosphere===============

Total Mass for top 15 floors  50700000 lbs 
Cross Section Area  43264 sf 
Drag Coefficient  1.28 
Altitude  1181 
Terminal Velocity  893 ft per second 
Terminal Velocity Calculator at NASA
http://exploration.grc.nasa.gov/education/rocket/termvr.html
At time t = s after being dropped,
the speed is v_{y} = m/s = ft/s ,
The distance from the starting point will be
y = m= ft The distance after being dropped is more than 1181 ft. , and will fall at the same rate as freefall
http://hyperphysics.phyastr.gsu.edu/hbase/traj.html
Freefall
At time t = s after being dropped,
the speed is v_{y} = m/s = ft/s ,
The distance from the starting point will be
y = m= ft
Of interest here is the fact that atmosphere alone would stop the top 15 Floors from accelerating just from drag, if you dropped it from above 12, 401 feet. This begs the question, how much would it be slowed down if it had to fall through the rest of the WTC tower?
The density of a WTC Tower is 371800000/59185152 = 6.28 lbs per cubic feet
Air density is 0.07 lbs per cubic foot
The density of the medium that a body falls through has an effect on it’s Terminal Velocity.
A body will fall in a liquid slower than it will through a gas, and even slower through something solid. When stuff gets in the way, it slows down the thing moving through it, the more stuff in the way, the slower the body goes, if it is not stopped entirely. A bullet goes through the air faster than it goes through a 10 foot thick brick wall, when you move something it takes work/energy to do it. In the case of things falling, we know exactly what that quanity of energy is. This refutes the “pancake theory” in a solid and provable way, based on the bedrocks of Newtonian physics. Things fall slower when they fall through denser medium.
========== A study of Grand Piano Fall Time in Atmosphere=============== The distance from the starting point will be
http://exploration.grc.nasa.gov/education/rocket/termvr.html Grand Piano Specifications: o Length: 161cm (5’3″) o Width: 149cm – o Height: 101cm o Weight: 285k o Finished in Black Polyester http://www.pianoplus.co.uk/yamahapiano/grands.html
4.88845 x 5.28215 = 25.8215261675 sq. ft.
Drag Coefficient = 1.28 (Flat Object CD 1.28 TV = 128.615 But those are the outside dimensions of the piano, pianos are not rectangular, but more like a right triangle, shaped in the most economical way to encase the stings. So they actual surface area could be more like 13 sf. rather than almost 26 sf. Then Terminal Velocity would be 181.889 ft/sec
At time t = s after being dropped,
http://exploration.grc.nasa.gov/education/rocket/termvr.html 
