Terminal Velocity: Solve Physics IA Problem

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Homework Help Overview

The discussion revolves around an investigation into terminal velocity, specifically examining how it varies with different masses of objects thrown from a height of approximately 23 meters. The original poster seeks clarification on the relationship between mass and terminal velocity squared, as well as the relevant equations involved.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the formula for terminal velocity and its components, questioning the correct interpretation of the slope in relation to mass and terminal velocity squared. There is discussion about the assignment of variables to axes in the graph.

Discussion Status

Some participants have provided insights into the formula for terminal velocity and its implications for the investigation. There is ongoing clarification regarding the correct interpretation of the slope and its relationship to the variables involved.

Contextual Notes

Participants are navigating potential misunderstandings about the formula and its application, as well as the implications of the mass of the objects being tested. The original poster is working within the constraints of an IB internal assessment.

Markus Lervik
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Hello,

I am finishing my IB internal assessment in physics. I have thrown four balls (with different masses, 400,450,475,500 grams) from a height, which is approximately 23 meters. My teacher told me to set up a graph which showed mass vs. vt^2.

He said that the inverse of M= ((1/2)(p)(C_d)(A) / (g)) x v^2 is the slope. I have problems making anything of this. Can someone please explain how to get up with the equation above?

Thank you very much!
 
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The purpose of the investigation was to see how terminal velocity acts on objects with different weight.
 
Moderator's note: moved to homework forum.
 
Markus Lervik said:
The purpose of the investigation was to see how terminal velocity acts on objects with different weight.

What is the formula for terminal velocity?
 
PeterDonis said:
What is the formula for terminal velocity?

v_terminal = (Sqrt(2mg)/(C*p_air*A))
 
Markus Lervik said:
He said that the inverse of M= ((1/2)(p)(C_d)(A) / (g)) x v^2 is the slope.
That is not the slope. The slope is either the ((1/2)(p)(C_d)(A) / (g)) part or its inverse. Which of those depends on how you assign mass and v2 to the x and y axes.
 

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