Question

Suppose that the terminal speed of a particular sky diver is 165 km/h in the spread-eagle position and 320 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 (Aslower / Afaster).

Answer 1

3.76.

Step-by-step explanation:

terminal velocity in spread eagle position = 165 km/h

terminal velocity in nose dive position = 320 km/h

ratio of cross section area = ?

terminal velocity of the diver is given by

`v = \sqrt{(2mg)/(C_(\rho)A)}`

A is the area of cross section

On rearranging

`A = (2mg)/(C_(\rho)v^2)`

From the above expression we can say that cross sectional area is inversely proportional to velocity.

`(A_(slower))/(A_(faster))=(v^2_(faster))/(v^2_(slower))`

`(A_(slower))/(A_(faster))=(320^2)/(165^2)`

`(A_(slower))/(A_(faster))=3.76`

Hence, the area ratio is 3.76.

Answer 2

3.76

Step-by-step explanation:

We are given that

Terminal speed in the spread -eagle position,
`v_t=165 km/h`

Terminal speed in the nosedive position,
`v'_t=320km/h`

We have to find the ratio of the effective cross-sectional area A in the slower position to that in the faster position.

We know that

Area, A=
`(2mg)/(C\rho v^2_t)`

`A_(slower)=(2mg)/(C\rho(165)^2)`

`A_(faster)=(2mg)/(C\rho(320)^2)`

`(A_(slower))/(A_(faster))=((320)^2)/((165)^2)`

`(A_(slower))/(A_(faster))=3.76`