Question about terminal speed using drag

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Homework Help Overview

The discussion revolves around calculating the terminal speed of a 75 kg man jumping out of a plane, modeled as a rectangular box. The problem involves understanding the appropriate cross-sectional area to use in the terminal velocity equation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the correct area to use in the terminal velocity equation, with some confusion about the dimensions and the concept of cross-sectional area. There are attempts to clarify the reasoning behind the area selection based on the orientation of the jump.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the correct area to consider, and there is acknowledgment of a potential resolution based on the feet-first position of the jumper.

Contextual Notes

Participants express frustration over discrepancies in calculations and the expected answer, indicating a need for clarity on the definitions and assumptions related to cross-sectional area in the context of terminal velocity.

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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