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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.

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  1. 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

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