How fast does a quarter go if you drop it off of the Empire State Building?

Answer 1

Do not try this, because it is dangerous to the people and property below.

The actual terminal velocity would be difficult to compute because there are multiple forces acting on a quarter as it falls: gravity, air resistance from it falling, and natural wind. The air resistance or drag depends on the density and viscosity of the air, which would vary with air pressure and humidity and temperature, to name a few variables. Also affecting air drag is the cross-section of the object falling, which for a coin will be changing as the coin tumbles through the air. I can tell you how to figure the gravity portion, though.

Two equations to describe an object falling in a vacuum (no wind, no air drag):

Vfinal = Vinitial + a*t and D = Vinitial * t + (a/2)*t2

V is velocity, t is time, a is acceleration, and D is distance. Taking Vinitial as zero, like you just dropped it and did not throw it, a = 9.8 m/s2, and D = 320 meters (see related link - there are observation decks at the 86th floor and 102nd floor. The website only mentioned an outdoor area at the 86th floor deck, which it said is 1050 feet (320 meters) high. Also it did not say the height of the 102nd floor, just that the entire building is 1454 feet tall, but I'm thinking this is the height of the entire building).

You don't know time, but can rearrange the two equations to get Vfinal = sqrt(2*a*D) = sqrt(2*(9.8 m/s2)*(320 m)) = 79.2 m/s, which works out to 177 miles per hour. (remember initial velocity assumed to be zero, and no wind or air resistance, though).

This is the maximum velocity an object could reach, but due to the other factors, it will be less than that. If for some way it was dropped from the 1454 feet (443.2 meters) then the maximum velocity at impact could be 93.2 m/s or 208.49 mph