Last question, please solve...?(posting again because no one answered)?

Question

The terminal speed of a sky diver is 160 km/h in the spread-eagle position and 315 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.


This is the answer:
3.8759765625


I just need to see how to get to the answer, I know it's simple, but I think I'm missing something THANKS :)

Answer 1

Air resistance causes a drag force that varies with velocity. The terminal velocity is the velocity at which the air resistance equals gravity, so there is no net force and hence the speed stays constant.

The formula for terminal velocity can be found at:
http://en.wikipedia.org/wiki/Terminal_velocity.

For your purposes, it can be summarized as:
Vterminal = K divided by the square root of A

where:
K is some combination of constants (including the drag coefficient) and

A is the cross-sectional area.

You know Vspread and Vnose so you can compute their ratio.
That gives you the ratio between the square roots of Aspread and Anose.

The ratio of square roots is just the square root of the ratio.

(A/B)^2 = (A/B)(A/B) = (AA)/(BB) = (A^2)/(B^2)

So that gives you your answer.

Answer 2

Hey nobody answered because they don`t know the answer. I don`t either. Good luck to you.