The terminal speed of a sky diver is 160 km/h in the spreadeagle position and 310 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.
Well of course C will change when the shape changes but anyway
Drag = c A v^2 = weight in both cases
C A1(160)^2 = C A2 (310)^2
A1/A2 = (310/160)^2 = 3.75