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

Answer:

3.76

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=$\frac{2mg}{C\rho v^2_t}$

$A_{slower}=\frac{2mg}{C\rho(165)^2}$

$A_{faster}=\frac{2mg}{C\rho(320)^2}$

$\frac{A_{slower}}{A_{faster}}=\frac{(320)^2}{(165)^2}$

$\frac{A_{slower}}{A_{faster}}=3.76$

Answer 2

Answer:

3.76.

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

$v = \sqrt{\dfrac{2mg}{C_{\rho}A}}$

A is the area of cross section

On rearranging

$A = \dfrac{2mg}{C_{\rho}v^2}$

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

$\dfrac{A_{slower}}{A_{faster}}=\dfrac{v^2_{faster}}{v^2_{slower}}$

$\dfrac{A_{slower}}{A_{faster}}=\dfrac{320^2}{165^2}$

$\dfrac{A_{slower}}{A_{faster}}=3.76$

Hence, the area ratio is 3.76.