Calculating Terminal Speed of a 82kg Skydiver using Basic Physics Formulas

AI Thread Summary
To calculate the terminal speed of an 82kg skydiver modeled as a rectangular box, the area (A) is determined to be 0.0924 m². The formula used is v = sqrt(2mg/(rho * C * A)), where the values for mass, gravitational acceleration, air density, and drag coefficient are substituted. The initial calculation yields a terminal speed of 135 m/s, but this result is questioned as incorrect. Discussions highlight the importance of the skydiver's orientation during the fall, suggesting that the largest surface area should be considered for accurate results. Ultimately, the consensus is to use the largest surface area of the box for the calculations.
Steelers72
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Homework Statement


A 82kg skydiver can be modeled as a rectangular "box" with dimensions 21cm× 44cm× 1.8m

Homework Equations


v = sqrt(2mg/(rho * C *A))

The Attempt at a Solution



A = 0.21 m * 0.44 m = 0.0924 m^2

v = sqrt(2 * 82 kg * 9.81 m/s^2 / (1.2 kg/m^2 * 0.0924 m^2 * 0.80))
=135m/sApparently this is wrong. I have tried many other formulae but I haven't had luck and I have one more attempt remaining. Any ideas?
 
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Why are you assuming that he dives "head first"?
 
Svein said:
Why are you assuming that he dives "head first"?

the question doesn't suggest otherwise, doesn't specify if he dives head first or legs first.
 
Steelers72 said:
the question doesn't suggest otherwise, doesn't specify if he dives head first or legs first.

What if the skydiver takes this attitude, before the parachute opens?

freefall2.jpg
 
SteamKing said:
What if the skydiver takes this attitude, before the parachute opens?

freefall2.jpg

Agreed! Take the largest surface of the box as area.
 
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