Finding cross-sectional area from terminal velocity.

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The discussion centers on estimating the cross-sectional area of a person based on the concept of terminal velocity and aerodynamic drag, which is proportional to the product of cross-sectional area and the square of velocity. The terminal velocity of a person is given as 56 m/s, but participants express confusion due to a lack of information regarding the person's mass and drag coefficient. It is noted that for terminal velocity, the drag force must equal the gravitational force, leading to the equation D = mg. Suggestions include using average human mass and the density of air, with a reference to the drag coefficient for convex bodies being approximately 1. The textbook answer for the cross-sectional area is 0.6 m², prompting further inquiry into the problem's parameters.
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Homework Statement



The aerodynamic drag on an object moving through air is proportional to Av^2, where A=cross-section, and v=velocity. The terminal velocity of a person without a parachute falling through the air is about 56m/s. Estimate the area of the cross-section of a person seen from the front.

Homework Equations


The Attempt at a Solution



I'm quite flummoxed by this one. We're told only that drag is propotional to Av^2. Anyway, I began by saying that for terminal velocity to be reached, the drag-force must equal the gravity-force, i.e. D = mg. But, I'm unsure how to proceed from there. I'm not told the mass of the person, nor the drag coefficient, so I really don't know how to find A. I'm sure the use of the word "estimate" here is significant. Maybe I'm expected to find the average human mass, and use that, but then that leaves me still without a complete equation, as I'm told the two are proportional.

The answer given in the textbook is 0.6m^2.

Help?
 
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You certainly have not been given enough information. As it happens, the drag coefficient for an arbitrary convex body tends to be about 1, but you still need to bring in the mass of the person and the density of the air.
I suppose you could relate the mass of the person to the cross-sectional area, assuming the person's density to be that of water, but that isn't quite straightforward.
This link might help: http://en.wikipedia.org/wiki/Drag_coefficient
 
Cheers, haruspex. I'll have to speak with my physics tutor and see if there's something I'm overlooking. Damn the holidays.
 

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