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

[newbe318]

Homework Statement

I attached a screen shot of the problem.

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.

 

[Simon Bridge]

Please show your best attempt with reasoning.

 

[newbe318]

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

 

[haruspex]

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?