Drag and Terminal Velocity

  • Thread starter dprimedx
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Information:
"In downhill speed skiing a skier is retarded by both the air drag force on the body and the kinetic frictional force on the skis. Suppose the slope angle is θ = 31.5°, the snow is dry snow with a coefficient of kinetic friction μk = 0.0400, the mass of the skier and equipment is m = 79.5 kg, the cross-sectional area of the (tucked) skier is A = 1.30 m2, the drag coefficient is C = 0.150, and the air density is 1.20 kg/m3."

Questions:
What is the terminal speed? Ok I got this part and came up with a velocity of 59 m/s.

"If a skier can vary C by a slight amount dC by adjusting, say, the hand positions, what is the corresponding variation in the terminal speed? (dvt/dC)"
This doesn't make any sense to me. I checked around online and all I can find is that dvt/dc=(-1/2)(k)(c^(-3/2)), but no one explains how they derived this. Also, the program I'm using seems to want an actual number for an answer but it doesn't tell me how much c is changing by. I also am unsure as to what 'k' is in the previous equation. Could someone explain what it is?

Can anyone help explain this to me? I'm not looking for the exact answer for the question, just an explanation as to how to arrive at it.
 
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Answers and Replies

  • #2
PhanthomJay
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I am assuming you got the first part right using the quadratic drag force formula to find the terminal velocity . If you differentiate it with respect to the drag coefficient, C, you'll get your answer, and discover what the constant 'k' is.
 

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