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

Question

A veritcal wind tunnel used to help train sky-divers has a range of air speeds as function of height within the main tunnel. At the bottom (y=0m) the air speed is 105 m/s and at the top (y=4m) the air speed is 25 m/s. At what height will a sky diver fly when tucked into a ball shape of radius .5 m, and when she maximizes her cross sectional area perpendicular to the air flow by spereaing out her body and limbs? Data: mass of sky diver= 100 kg, drag coefficients Dball =.5 and Dbody=1.2 density of air= 1.23 kg/m3, cross sectional areas: .445m2 and 1.25m2 when in ball or spread out respectively

Explanation / Answer

Here,

for the speed of wind as a function of h

v = 105 - b * y

at y = 4 , v = 25

25 = 105 - b * 4

b = 20 1/s

v = 105 - 20 y m/s

Now , when she is tucked in a sphere

Drag force = gravity

m * g = 0.5 * Dball * p * v^2 * A

100 * 9.8 = 0.5 * 0.5 * 1.23 * (105 - 20 y )^2 * 0.445

solving for y

y = 1.019 m

the height of diver when tucked into a ball shape is 1.019 m

--------------------------------------

when she maximizes the cross sectional area

Drag force = gravity

m * g = 0.5 * Dbody * p * v^2 * A

100 * 9.8 = 0.5 * 01.2 * 1.23 * (105 - 20 y )^2 * 1.25

solving

y = 3.62 m

the height of diver will be 3.62 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects