As I have mentioned before, I have argued with many “truthers,” one of whom I have done a couple of articles on- we’ll call him “truther” A. One of his buddies, who I will call “truther” B, is just as idiotic.

For example, “truther” B is the one who insists that you can’t move more with less and when we pointed out that a person can push a car, his lame excuse was that the car has wheels. The fact that the car has wheels does NOT change the fact that a car has more mass than a person.

Another thing “truther” B did is accuse me of saying the steel took on the properties of air when the WTC collapsed when I said no such thing.

I merely stated that Brent Blanchard correctly stated that the WTC, like most human-occupied buildings, is mostly air and that air is a gas ( well, technically, a mixture of gases, but you get the picture). Brent Blanchard also states that in a collapse such as what occurred in the WTC, air has to be displaced in some manner.

What I did was state the physics law that explains why the air was displaced, which was Boyles Law, pertaining to the behavior of gases. Boyle’s Law, as any physics book will tell you, states that the volume of air in a container is inversely proportional to the pressure. This means that if the pressure increases, the volume decreases and vice versa. It is often written as P_{1}V_{1 } = P_{2}V_{2}. I simply pointed out that the WTC ( or any human-occupied building in general) can be considered a container, and since the Twin Towers were collapsing, the volume was obviously decreasing, which means that, per Boyle’s law, the pressure had to increase. In short, the increasing pressure created by the collapsing floors shoved the air out; the collapsing floors behaved much like the plunger in a syringe.

Now, if you are wondering why the debris was curving as it fell, as you see from the video here, that, again is explained by physics. The equations I used in my previous article may be used for horizontal motion as well as vertical motion- oh, and horizontal and vertical motion are INDEPENDENT from each other. To recap, the equations are as follows: To find the distance something goes, the equation, as found in any physics book, is d = v_{i}t + ½ at^{2 } where d is the distance, v_{i} is the initial velocity, a is the acceleration ( if you are referring to horizontal motion, a is acceleration meaning change in velocity over time, if you are referring to vertical motion, a is acceleration due to gravity), and to find the final velocity, the equation is v = v_{i }+ at, where v is the final velocity, v_{i }is the initial velocity, a is acceleration, and t is time.

Let’s say that a chunk of debris was shoved out from about 800 ft ( 244 meters). In both cases, the initial velocity is zero, so we only need the 2^{nd} half of each equation. Obviously, if the debris is 244 meters up, then we know the vertical distance it falls, and only need to find the time- again, time will be the same due to the fact that horizontal and vertical motion are INDEPENDENT from each other. So, for the first equation, we only need to use d = ½ at^{2 } to figure out how long it takes the debris to hit the ground- in this case, the acceleration is the acceleration due to gravity. Rearrange the equation appropriately and plug in the numbers and we find it takes the debris about 7 seconds to hit the bottom.

Now, say its horizontal velocity when ejected was about 5 m/s ( acceleration would be 5 m/s^{2}). Since the initial horizontal velocity was zero, we can find the horizontal distance traveled by using this equation: d = ½ at^{2} = ½ (5m/s^{2})(7 s)^{2 }= 123 meters or just over 400 ft.

These may also be used if the initial velocity is greater than 0. For example, if you throw a ball straight down off a cliff with an initial velocity of 5 m/s and it takes the ball 5 seconds to reach the ground, you need to use the FULL equation d = v_{i}t + ½ at^{2}^{ }= (5 m/s) (5 s) + ½ (9.8m/s^{2}) ( 5s)^{2} = 148 meters or 483 ft. I used the acceleration due to gravity because once that ball leaves your hand, only gravity is acting on it.

These equations may also be used for both horizontal and vertical motion. For example, say someone is driving a car at about 65 MPH (29 m/s), approaching a 100 ft ( 30 meter)-high cliff, and instead of turning, suddenly accelerates to 75 MPH ( 34 m/s) and drives off the cliff ( which, of course, I would NOT recommend). The initial vertical velocity is 0, so we can figure out how fast the car would take to hit bottom by rearranging the equation d = ½ at^{2 } appropriately and find that it takes the car about 2.5 seconds to hit the bottom.

We can also see how far the car travels horizontally- in this case, the acceleration would be 34 m/s – 29 m/s divided by 2.5 seconds = 2 m/s^{2 }. To find the horizontal distance, we use the full distance equation, so d = 29 m/s ( 2.5s) + ½ (2m/s^{2})( 2.5s)^{2} = 78.75 meters or about 258 ft.

In both the case of the car, and the debris, before they go over the edge or out the building, respectively, the ground ( or the floor) are holding them in place. However, once the ground ( or the floor) is removed, gravity kicks in, resulting in a downward motion. However, since both the car and the debris also have horizontal motion when going off the cliff or being ejected by the pressure of a collapsing building, the combination of the horizontal and vertical motion results in a curve, unless you’re using Roadrunner cartoon physics.

Now, getting back to the couch that “truther” A said someone could ride down from the upper floors and survive in my previous article, as you may remember, the couch took 8 seconds to hit bottom. We can use the equation v = v_{i} + at to figure out how fast the couch is going at each second until impact. Again, since the initial velocity is 0, we can simply say v = at. Acceleration would be 9.8 m/s^{2}, and the information is in the following table:

Time in seconds (s) | Acceleration | Velocity in meters/second | Velocity in mph |

1s | 9.8 m/s^{2} |
9.8 m/s | 22 mph |

2s | 9.8 m/s^{2} |
19.6 m/s | 44 mph |

3s | 9.8 m/s^{2} |
29.4 m/s | 66 mph |

4s | 9.8 m/s^{2} |
39.2 m/s | 88 mph |

5s | 9.8 m/s^{2} |
49 m/s | 110mph |

6s | 9.8 m/s^{2} |
58.8 m/s | 132 mph |

7s | 9.8 m/s^{2} |
68.6 m/s | 153 mph |

8s | 9.8 m/s^{2} |
78.4 m/s | 175 mph |

We can also find out the kinetic energy of the couch at each second until impact- as any physics book will tell you, kinetic energy is equal to ½ the mass times the velocity SQUARED, written as KE = ½ mv^{2}. As mentioned in my previous article, the mass of the couch was between 80-90 pounds or about 36-41 kilograms. To make the math easier, let’s say the couch weighs 85 lbs, which is about 39 kilograms, so the following table indicates the kinetic energy of the couch at each second:

Time (s) | Massin kg | VelocityIn m/s | Velocity squared | Kinetic energy in joules(J) | |

1 | 39 | 9.8 | 96.04 | 1873 | |

2 | 39 | 19.6 | 384.16 | 7491 | |

3 | 39 | 29.4 | 864.36 | 16,855 | |

4 | 39 | 39.2 | 1536.64 | 29,964 | |

5 | 39 | 49 | 2401 | 46,820 | |

6 | 39 | 58.8 | 3457.44 | 67,420 | |

7 | 39 | 68.6 | 4705.96 | 91,766 | |

8 | 39 | 78.4 | 6146.56 | 119,858 | |

As you can see, that couch hit bottom with a kinetic energy of just under 120,000 joules- it’s highly unlikely the couch would remain intact when it hit bottom and neither would anyone on the couch. Clearly, the “truthers” need to lay off the kool-aid.