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!