Question

The terminal speed of a sky diver is 130 km/h in the spread-eagle position and 326 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.

Answer 1

Answer:

$\frac{A_{slow} }{A_{fast} }=6.2885$

Explanation:

Given data

Terminal velocity for spread eagle position vt=130 km/h

Terminal velocity for nosedive position vt=326 km/h

The terminal speed of the diver is given by

$v_{t}=\sqrt{\frac{2mg}{CpA} }$

Therefore the area is given by

$A=\frac{2mg}{Cp(v_{t})^{2} }$

Since everything else is constant in the two dives except for the terminal velocity, the ratio between the area in the slow position to the area in the fast position is

$\frac{A_{slow} }{A_{fast} }=\frac{326^{2} }{130^{2} }\\\frac{A_{slow} }{A_{fast} }=6.2885$