What is the Terminal Velocity of a Skydiver with Given Parameters?

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SUMMARY

The terminal velocity of a skydiver can be calculated using the equation D = 1/2 C p A v^2, where D is the drag force, C is the drag coefficient, p is the air density, A is the cross-sectional area, and v is the velocity. In this discussion, the parameters provided include a mass (m) of 84 kg, a drag coefficient (C) of 0.8, an air density (p) of 1.2 kg/m^3, and a cross-sectional area (A) of 987 cm^2. The correct formula to find terminal velocity is v = sqrt((2mg)/(CpA)), which requires converting the area from cm^2 to m^2. The initial calculation of 1.38 m/s is incorrect due to the area conversion oversight.

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newbe318
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Homework Statement


I attached a screen shot of the problem.
Screen Shot 2015-10-14 at 9.40.16 PM.png


Homework Equations


D= 1/2 C p A v^2
bv=mg

The Attempt at a Solution


I spent 3 long hours on this problem. I confused myself even more. Please help me understand what is going on.
 
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Please show your best attempt with reasoning.
 
I know that at terminal velocity, bv=mg ... so, mg=1/2CpAv^2.. would I just solve for v? .. v=sqrt((2mg)/(CpA))... v=1.38m/s ?
m: 84kg
C: 0.8
p: 1.2kgm^3
A: 987cm^2
 
newbe318 said:
I know that at terminal velocity, bv=mg ... so, mg=1/2CpAv^2.. would I just solve for v? .. v=sqrt((2mg)/(CpA))... v=1.38m/s ?
m: 84kg
C: 0.8
p: 1.2kgm^3
A: 987cm^2
That's the right approach, but I don't understand how you get such a small velocity. Did you forget to convert the area to m2?
 

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