Here are some more real world physics for “truthers.”

Let’s say you have a penny that was made after 1982, which has a mass of 2.5 grams or 0.0025 kilograms (Source). Yes, I know pennies made before 1982 have a different mass; I’m just sticking to post-1982 pennies to simplify matters. If you drop that penny on your foot, it wont hurt.

However, if you drop the penny off the 102nd floor of the Empire State Building, which is a height of 1,224 feet or 373 meters (source), if it hits anyone, that penny WILL hurt as it will acheive a rather impressive velocity.

Here are some calculations to show how far the penny will fall after certain time intervals, it’s velocity at those intervals, and it’s kinetic energy. The equations used come from my old physics book I still have, __The Physics of Everyday Phenomena,__ Fourth edition, by W. Thomas Griffith.

First, we start with seeing how far it will fall once you release the penny: The equation for this is d = v-naught (time) = ½ at². Since the initial velocity in this case is 0, we can also write this as d = ½ at². If you want to see how long it takes the penny to hit the ground, simply rearrange the equation accordingly and you’ll see the penny takes 60 seconds to hit the ground. Remember, a is acceleration due to gravity, which is 9.8 m/s² for the earth, so here’s the table for that (I’m using the metric system but you can go here to convert to English units):

Time (s) |
½ at² = |
Distance in meters |

1 | ½ (9.8m/s²) (1s)² = | 5 m |

2 | ½ (9.8m/s²) (2s)² = | 20 m |

3 | ½ (9.8 m/s²)(3s)² = | 44 m |

4 | ½ (9.8m/s²)(4s)² = | 78 m |

5 | ½ (9.8 m/s²)(5s)² = | 123 m |

60s | ½ (9.8m/s²)(60s)² | 373 m |

Now, for the velocity, we use this equation, v = v-naught + at, but, again, as v-naught is 0, we can simply write v = at: (remember, 1 m/s is 2.24 MPH):

Time (s) |
a(t) = |
Velocity in meters/second |
Velocity in MPH |

1s | 1s(9.8m/s²) | 9.8 m/s | 22 MPH |

2s | 2s(9.8m/s²) | 20 m/s | 44 MPH |

3s | 3s(9.8 m/s²) | 29 m/s | 66 MPH |

4s | 4s( 9.8m/s²) | 39m/s | 88 MPH |

5 s | 5s(9.8m/s²) | 49m/s | 110MPH |

60s | 60s(9.8m/s²) | 588m/s | 1315 MPH |

As you can see, that penny’s moving pretty fast when it hits the ground. Now, here’s the kinetic energy, KE = ½ mv²

Time |
½ mv² = |
Kinetic energy in Joules |

1s | ½(0.0025 kg)(9.8m/s)² | 0.12 Joules |

2s | ½(0.0025 kg)(20m/s)² | 0.48 Joules |

3s | ½(0.0025 kg)(29m/s)² | 1.1 Joules |

4s | ½( 0.0025kg)(39m/s)² | 2.0 Joules |

5s | ½(0.0025kg)(49m/s)² | 3.0 Joules |

60s | ½(0.0025kg)(588m/s)² | 432 Joules |

That’s a lot of energy for a tiny little penny, even accounting for air resistance. Before any “truther” actually does the calculations and says they don’t quite match, it’s called rounding off.

Now, if a little penny can get going that fast and have that much energy, when it’s mass is constant, imagine what the collapsing upper 16 floors of the north tower, which weighed over 435,000 tons or 394,625,361 kilograms would do to the floors below them. As Ryan Mackey points out on page 44 of “On Debunking 911 Debunking,” yes, the collapsing floors would decelerate briefly when hitting the floors below them, but that deceleration’s negligible.

Before some “truther” points out the obvious fact that the penny is not the collapsing Twin Towers, I KNOW that; that’s not the point. The point is that if a penny, which has a fairly low mass, can move that fast, then the much more massive 16 floors of the North Tower, which would have an increasing mass (When those 16 floors hit the floor below them, you must factor in the mass of the new floor, and so on) and velocity, would have a very devestating effect and it’s obvious to anyone that understands physics that there’s no way the Twin Towers could have remained standing.