Question about terminal speed using drag

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The terminal speed of a 75 kg man jumping out of a plane feet first can be calculated using the equation v terminal = sqrt(4mg/(p * area)). The correct cross-sectional area to use in this scenario is 0.08 m², derived from the dimensions 0.20 m (width) and 0.40 m (height), since he is oriented feet first. This results in a terminal velocity of approximately 170 m/s. The initial confusion stemmed from using the incorrect area of 0.72 m², which does not account for the jump position. Understanding the correct orientation and area is crucial for accurate calculations.
sona1177
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What is the terminal speed of a 75 kg man that jumps out of a plane feet first? The man can be thought of as a rectangular box with dimensions 20 cm * 40 cm * 1.8 m

I tried equation v terminal = sq rt (4mg/p* area). P=1.22
I used . 72 as Area. That's wrong. Help please!

I calculated eq using .72 but that is wrong. Help!
 
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What area should be used in the equation for terminal velocity?

ehild
 
See my book says to use the cross sectional area. The answer is supposed to be 170 m/s but I am very frustrated because I can't see how they got that. His cross sectional area is 1.8 m * .40m =.72 but it's wrong if you use this.
 
Oh i see what you are saying, since he jumps out feet first, that means the area is .20 * .40, this side meets air resistance. If you use this in the eq the value is 170 m/s and that's the answer. Is my reasoning correct?
 
sona1177 said:
Oh i see what you are saying, since he jumps out feet first, that means the area is .20 * .40, this side meets air resistance. If you use this in the eq the value is 170 m/s and that's the answer. Is my reasoning correct?

Yes, that is the clue, feet first! Good job!

ehild
 
Thanks for all your help!
 

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