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

`(A_1)/(A_2)=3.97549776055`

Step-by-step explanation:

The equation for terminal velocity is
`v_t=\sqrt{(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=(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=(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
`(A_1)/(A_2)=(v_(t2)^2)/(v_(t1)^2)=3.97549776055`