Question

The terminal speed of a sky diver is 163 km/h in the spread-eagle position and 325 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:

$\frac{A_1}{A_2}=3.97549776055$

Explanation:

The equation for terminal velocity is $v_t=\sqrt{\frac{2mg}{\rho AC_d}}$, where m is the mass of the sky diver, g the gravitational acceleration, $\rho$ the air density, A the effective cross-sectional area and $C_d$ the drag coefficient.

It will be easier if we write this as $v_t^2 A=\frac{2mg}{\rho C_d}$ and realize that for both situations the right hand side of that formula will be the same. This means that $v_{t1}^2 A_1=v_{t2}^2 A_2=\frac{2mg}{\rho C_d}$, so the ratio of the effective cross-sectional area A in the slower position ($A_1$ for $v_{t1}=163 km/h$) to that in the faster position ($A_2$ for $v_{t2}=325 km/h$) will be $\frac{A_1}{A_2}=\frac{v_{t2}^2}{v_{t1}^2}=3.97549776055$