What Is the Drag Force on a Sky Diver at Different Speeds?

AI Thread Summary
The discussion focuses on calculating the drag force on a sky diver at different speeds. The diver, with a mass of 79.5 kg, reaches a terminal speed of 53.0 m/s, with an acceleration of 6.792 m/s² at 30.0 m/s. The drag force at terminal velocity is calculated to be 779.1 N, but the user struggles with finding the drag force at 30.0 m/s. It is noted that drag force is proportional to the square of velocity, leading to the suggestion to derive the drag coefficient from the terminal velocity to solve for the drag force at the lower speed. The conversation also touches on the importance of significant figures in the calculations.
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Homework Statement



A sky diver of mass 79.5 kg jumps from a slow-moving aircraft and reaches a terminal speed of 53.0 m/s.

(a) What is the acceleration of the sky diver when her speed is 30.0 m/s?

What is the drag force on the diver when her speed is the following?
(b) 53.0 m/s (answer in N)
(c) 30.0 m/s (answer in N)

Homework Equations


The Attempt at a Solution



I found parts a and b (correct on Webassign), but I can't get part c.):

a.) a= 6.792 m/s2
b.) 779.1 N

Would part c.) just be: mg-ma?

(779.1)-(79.5)(6.792)
= 239.096 (WebAssign says it's incorrect)
 
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Drag force in air ia approximately proportional to v^2. So f = D*v^2.
Using the result from b, find D.
Use this value to solve c.
 
Thanks. I got it.
 
Your answer looks good. Does WebAssign concern itself with doing significant figures properly?
 

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