Calculate terminal velocity when given cross sectional area

[Tatortotts]

Homework Statement

Two packages are dropped from an airplane. A parachute can increase the cross sectional area of each packages by a factor of 31. The parachute on package 1 fails to open, and the terminal speed of package 1 is 10 m/s. The parachute on package 2 opens.

What is the terminal speed of package 2 in m/s?

Homework Equations

Unsure, considering the problem gave very few numbers to work with.
R = ½CρAv²

The Attempt at a Solution

I figured since A (cross sectional area) is inversely proportional to v_terminal², that you would square package 1's v_terminal (to get 100 m/s) and then divide by 31. I haven't submitted my answer yet so I'm not sure if it's right.

 

[haruspex]

I figured since A (cross sectional area) is inversely proportional to vterminal2, that you would square package 1's vterminal (to get 100 m/s) and then divide by 31.

Have you left out a final step?

 

[Tatortotts]

Have you left out a final step?

Well, v_terminal² is proportional to 1/A, so 10² = 1/(A*31)
Solving for A I then get 1/3100 m/s, which seems a little ridiculous.
I submitted my answer (from my original solution--3.2 m/s) and it was wrong, so I know I'm missing something. I just can't seem to figure out what.

 

[haruspex]

Well, v_terminal² is proportional to 1/A, so 10² = 1/(A*31)
Solving for A I then get 1/3100 m/s, which seems a little ridiculous.
I submitted my answer (from my original solution--3.2 m/s) and it was wrong, so I know I'm missing something. I just can't seem to figure out what.

Here's a hint. The square of 10m/s is not 100m/s. What is it?

 

[Tatortotts]

Here's a hint. The square of 10m/s is not 100m/s. What is it?

Would it then be 100m²/s²? So I should take the square root of my answer?

 

[haruspex]

Would it then be 100m²/s²? So I should take the square root of my answer?

Yes.

 

[Tatortotts]

Yes.

Thank you so much! I don't know why I decided to neglect units for that problem.