Question

The terminal speed of a skydiver is 160 km/h in the spread eagle position and 310 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:

$\dfrac{A_{slow}}{A_{fast}}=3.753$

Explanation:

given,

terminal speed of skydiver = 160 km/h                      

At nose dive position = 310 km/h                                            

using the equation of terminal velocity                                

$A = \dfrac{2mg}{C_pV^2_r}$                                        

$\dfrac{A_{slow}}{A_{fast}}=\dfrac{V^2_{fast}}{V^2_{slow}}$

$\dfrac{A_{slow}}{A_{fast}}=\dfrac{310^2}{160^2}$            

$\dfrac{A_{slow}}{A_{fast}}=3.753$                      

ratio of the effective cross sectional area A = 3.753