How to determine air-resitance for a dropped coin

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To determine air resistance for a coin dropped from 2 meters, the terminal velocity equation can be used, referencing experiments like those from Mythbusters that measured drag coefficients. However, air resistance has a minimal impact in this scenario, as the coin reaches the ground in approximately 0.67 seconds at a speed of 6.5 m/s, while its terminal velocity is around 40 m/s. The coin's tumbling motion complicates measurements, as the orientation affects drag; the flat side facing down will experience more resistance than the edge. Ultimately, the effect of air resistance in this case is negligible. Understanding these dynamics can provide insight into the physics of falling objects.
16physicist
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Can anyone tell me how I can determine air-resitance for a coin being dropped from 2 meters onto a concrete floor?

Thanks a lot.

-Arash Ay
 
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Depends on what, exactly, you are looking do figure out. Mythbusters did an episode on this, iirc, where they actually measured the terminal velocity, so you could take that and plug it into the terminal velocity equation to find the drag coefficient.

Realistically, though, when dropped from 2 meters, drag will play an insignificant role in what happens. The coin hits the ground in about 2/3 of a second, traveling at 6.5 m/s. The terminal velocity, on the other hand, is probably somwhere around 40 m/s.
 
This would be very difficult because when the coin is dropped it will flip over and over so the same surface will not always be facing the ground. When the flat side is down the resistance will be more than when the thin edge is facing down. But like Russ said it would be so small and would not make much difference.
 
Okay, I apperciate that, thank you.
 

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