001 Terminal

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Can a coin/any simmilarly sized object kill you if it is dropped from a skyscraper/high point onto your head?

If it is thrown with enough spin so that it stays on it's side, not only will it have a lot less friction, but the point of impact being much smaller, will create a larger pressure.

A penny is
m = 2.5g = 0.0025kg
thick = 1.55mm = 0.00155m
width = 19mm = 0.019m
Cross section Area = 0.00155m*0.019m = 0.00002945m^2
Drag coefficient = 0.05 (between a smooth sphere = 0.4 and a laminar flat plate = 0.001)

Terminal velocity is
Vt = sqrt(2mg/ρACd) where
m is mass = 0.003kg
ρ is density of air = 1.2kg/m^3
A is area
Cd is drag coefficient estimated at 0.05, somewhere between a smooth sphere 0.4 and a laminar flat plate 0.001

Vt = sqrt(2*0.0025*9.8/1.2*0.00002945*0.01)
Vt = 166 m/s

The Cd is crucial. If flat, the Cd = 1.18 and Vt = 34m/s

The skin and skull on your head has a thickness of half a centimeter. So assuming the penny punctures the skull, it will go from 166m/s to 0 over a distance of half a centimeter.

This represent an acceleration of
a = 1/2 (166)^2/0.005 = 2 773 000 m/s^2

F = ma = 2 773 000m/s^2*0.003kg = 8320 N

or a force of 832 kg ...

This force will be distributed over a maximum surface of 29mm^2 (1.5 mm thick by 19 mm wide). This is probably ten time too big since the cross section is not flat, but curved:

So the pressure is

P = F/A = 8320N/(0.00002945m^2) = 282 485 061 Pa
P = 282 MPa = 40 900 PSI

Looking up the ultimate compressive strength of some common materials,
(ksi=1000psi)

Oak 7ksi
Granite 15-26 ksi
Steel 30-60 ksi

The penny would puncture the skull.

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