Finding ratio of smaller terminal velocity to larger terminal velocity

[rockchalk1312]

The terminal speed of a sky diver is 188 km/h in the spread-eagle position and 320 km/h in the nosedive position. Assuming that the diver's drag coefficient C does not change from one position to the other, find the ratio of the effective cross-sectional area A in the slower position to that in the faster position.


vt = sqroot (2mg/pAC)


188=sqroot(2(90)(9.8)/(1)(A)(1))

320=sqroot(2(90)(9.8)/(1)(A)(1))

Basically I just set v to each value given, set everything else the same and solved for A then divided the smaller value by the larger one and got .34 but this was the wrong answer.

Really thought this was a pretty simple thing to figure out but I'm probably just making a stupid mistake? Thank you!

 

[LawrenceC]

"find the ratio of the effective cross-sectional area A in the slower position to that in the faster position."

Should the ratio be greater than or less than unity?

 

[rockchalk1312]

I'm sorry, what do you mean by unity?

 

[LawrenceC]

Unity means the number one. Should the ratio be greater than or less than one?

 

[rockchalk1312]

Unity means the number one. Should the ratio be greater than or less than one?

Ok yeah I see it should be greater than 1 since the area of A for the slower one would be greater than for the faster one...but was it still ok to approach the problem the way I did?

 

[rockchalk1312]

Ok it was just the reciprocal! Thank you very much :)

 

[LawrenceC]

Glad you got it. I was trying to give you a subtle hint rather than saying take the reciprocal.