The terminal speed of a sky diver is 132 km/h in the spread-eagle position and 304 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.

5.304

Explanation:

The terminal speed of the diver is given by vt = (2mg/cpa) 1/2

Therefore, the area is given by A = 2mg/cpvt2

Since every thing else is constant, in the two dive except the terminal velocity, the ratio between the area ion the slow position to the area in the fast position is A slow/A fast

this gives 304square/132square = 5.304