Speed and Velocity
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).
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Update: June 2012