1) A question from lecture: can/would a penny dropped from the top of the Empire

Question

1) A question from lecture: can/would a penny dropped from the top of the Empire State Building kill a pedestrian at
ground level? This a very old and question that has persisted in American culture for generations since the thrill of
the days the this building reigned as the world’s tallest. Trivial answer: No, because the building tapers upward so a
penny dropped from the Observation Deck would probably just land on some part of the building below. But let’s
play physicist and reduce the problem to evaluate the potential outcome and assume that you could drop a penny
from the height of the Observation Deck on the 102nd floor (1,250 ft, 381 m).
Modern pennies weigh 2.5 grams, but prior to 1982 the US Mint struck pennies in pure copper that weighed 3.11
grams. Use the pre-1982 weight to best address the original notion of killer pennies. Pennies have a diameter of
0.750 in (19.1 mm) and a thickness of 1.52 mm.
a) A falling penny likely tumbles. Calculate the maximum and minimum terminal velocities that a penny could achieve
at different orientations to get some boundaries on the problem, expecting the actual velocity to fall somewhere
between the two extremes.

b) Using the maximum terminal velocity that you found, calculate the energy a penny would impart on impact with
your head, and interpret the potential lethality of such an impact?

Explanation / Answer

a.) Vmax = sqrt(2mg/rho*A*C)

C is drag coefficient

A is area

rho is density of air

For Vmax A= 2*pi*r*h

Vmax = sqrt(2*3.11*10^-3*9.81/2*pi*((19.1*10^-3)/2)*1.52*10^-3)

Vmax= 25.86 m/s

for Vmin A= pi*r^2

Vmin = sqrt(2*3.11*10^-3*9.81/pi*((19.1*10^-3)/2)^2)

Vmin= 14.59 m/s

b.)energy stored in penny = 0.5mv^2

E= 0.5*3.11*10^-3*(25.86)^2 = 1.039J

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