A vertical wind tunnel used to help train sky-divers has a range of air speeds a

Question

A vertical wind tunnel used to help train sky-divers has a range of air speeds as a function of height within the main tunnel. At the bottom m 00.0 y the air speed is 105 m/s and at the top m 4.00 y , the air speed is 25.0 m/s. At what height will a sky-diver “fly” when (a) tucked into a ball shape of radius 0.500 m and (b) when she maximizes her cross-sectional area perpendicular to the air flow by spreading out her body and limbs? Data: mass of sky-diver = 100 kg, drag coefficients 500.0 ballD and 20.1 bodyD , density of air =1.23 kg/m3, cross-sectional areas: 0.445 m2 and 1.25 m2 when in ball or spread-out, respectively.

Explanation / Answer

Drag force on sphere = Cd*(rho)v^2*(area)
Cd = 500
[v(h) - v(H)]/[h - H] = [v(0) - v(H)]/[-H]
(v - 25)/(h - 4) = (105 - 25)/(-4)
v - 25 = (h - 4)*(-20) = -20h + 80
v = 105 - 20h

D =500*1.23*0.445*(105 - 20h)^2/2 = 100*9.8 = mg
105 - 20h = +- 2.67
h = 5.11m, 5.38m

for open body

D = 20.1*1.23*1.25*(105 - 20h)^2/2 = 9.8*100
105 - 20h = +-7.96
h = 4.952m , 5.648

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