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Name: Mike
Status: educator
Age: 40s
Location: N/A
Country: N/A
Date: 1999 

How do you calculate the speed and velocity of a penny that is droped from the top of the Empire State building?

First we must find the terminal velocity, to see if the thing reaches it before it reaches the ground. We set the aerodynamic drag equal to the force of gravity:

F_drag = 0.5 c p v^2 A
   c = coefficient of drag for a thin airfoil = 0.05 (or so)
   p = density of air = 1 g/L
   v = terminal velocity of penny = ?
   A = frontal area of falling (edge-on, we assume) penny,
        about 18 mm x 1 mm

F_gravity = m g

   m = mass of penny = 2 g (a guess)
   g = 9.8 m/s^2

   v = Sqrt[(m g)/(0.5 c p A)] = 200 m/s 
( >1 sig fig not justified )

Penny would reach this velocity in 200 m/s / 9.8 m/s^2 = 20 seconds, and it would travel (from rest) 0.5 9.8 20^2 = 2,000 m in that time. The Empire State isn't 2 km high, so we conclude the penny will not reach terminal velocity.

Hence the velocity is found roughly by using
    h = 0.5 g t^2 = height of building = 391 m to observation deck
    v = g t

    Gives v = g Sqrt[h/(0.5 g)] = 88 m/s = 200 MPH. 

Since drag will be strong already at this speed, this must be an upper estimate and we expect the actual velocity to be lower.


Well, you really need to ignore air resistance here. In reality, a penny falling sich a distance through the air will begin to tumble, making its path chaotic. When that happens, you really can't know in advance exactly how fast it will be falling when it lands.

The easy way to calculate this is to first calculate the potential energy of the penny at the top of the building (relative to its potential energy at the sidewalk), then assume that when it reaches the sidewalk, all its former potential energy is converted to kinetic energy. If the height if the building is H and the mass of the penny is M, the potential energy of the penny at the beginning is MgH, where g is the acceleration due to gravity, 9.8 m/s^2 (or, more usefully for this problem, 9.8 N/kg). At the end of the fall, the penny's kinetic energy is MgH = 0.5 M v^2, where v is the magnitude of the velocity. You can solve this equation for v:
MgH = (0.5) M v^2
gH = (0.5) v^2
2gH = v^2
v = (2gH)^(0.5)

In other words, the velocity will be the square root of the quantity 2gH.

Richard Barrans Jr., Ph.D.


Method to use depends on the accuracy that is desired.

You can use a stop watch. Shortly after the penny is dropped, it will reach a steady state terminal velocity. You can measure the time between when it is dropped (one needs sharp eyes or a telescope) and when it hits the ground. I suspect it would take about 20 second or so, depending on wind, etc. If, as I am assuming, the penny reaches its "terminal velocity" rather early in its fall, then you could expect an accurate measured average velocity simply by using a stop watch.


Dr. Ali Khounsary
Advanced Photon Source

To do the problem completely you'd need to know the drag force on a penny as a function of its speed through air. This would allow you to write a differential equation for the penny's acceleration. However, it's a pretty safe bet that the penny will reach terminal velocity before it lands, so if you don't care about how long it takes to fall, or how its speed varies as it falls, you'd just need that one number, which I have no idea how to calculate.

There's an element of randomness here, of course, because the drag and the path the penny takes depend on the penny's orientation.

You could get a quick upper limit of the penny's speed by ignoring the drag, calculating the total change in potential energy m*g*h, and assuming it all gets converted into kinetic energy (1/2)m*v^2. Then v = sqrt(2*g*h).

Tim Mooney

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