Question

uppose that the terminal speed of a particular sky diver is 150 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

4.55

Step-by-step explanation:

The terminal speed of a diver is given by:

`v_t=\sqrt{(2mg)/(C\rho A) } \\\\Where\ m=mass\ of \ driver,d=acceleration\ due\ to\ gravity,C=drag\ \\coefficient,A=cross\ sectional\ Area.\\\\Therefore:\\\\A=(2mg)/(C \rho v_t^2) \\\\For\ area\ with\ terminal\ speed\ in\ spread\ angle\ position(v_s):\\\\A_s=(2mg)/(C \rho v_s^2) \\\\For\ area\ with\ terminal\ speed\ in\ nose\ dive\ position(v_n):\\\\A_n=(2mg)/(C \rho v_n^2)\\\\Therefore\ since\ g,m,C,\rho\ are\ constant:\\\\`

`(A_s)/(A_n)= ((2mg)/(C \rho v_s^2))/((2mg)/(C \rho v_n^2))\\\\(A_s)/(A_n)= (v_n)/(v_s) \\\\v_n=320\ km/h,v_s=150\ km/h\\\\(A_s)/(A_n)=(320^2)/(150^2) =4.55`