Terminal Velocity: Solve Physics IA Problem

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The discussion focuses on an IB physics internal assessment investigating terminal velocity by analyzing the fall of four balls of varying masses from a height of 23 meters. The teacher suggested creating a graph of mass versus terminal velocity squared (vt^2) and indicated that the slope relates to the equation M = ((1/2)(p)(C_d)(A) / (g)) x v^2. However, there is confusion regarding the correct interpretation of the slope, which could either be the constant ((1/2)(p)(C_d)(A) / (g)) or its inverse, depending on the axes chosen for the graph. Clarification is sought on how to apply the equation correctly in the context of the experiment. Understanding these relationships is crucial for accurately representing the effects of mass on terminal velocity.
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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