# Terminal velocity

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1. Mar 18, 2017

### Markus Lervik

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!

2. Mar 18, 2017

### Markus Lervik

The purpose of the investigation was to see how terminal velocity acts on objects with different weight.

3. Mar 18, 2017

### Staff: Mentor

Moderator's note: moved to homework forum.

4. Mar 18, 2017

### Staff: Mentor

What is the formula for terminal velocity?

5. Mar 18, 2017

### Markus Lervik

v_terminal = (Sqrt(2mg)/(C*p_air*A))

6. Mar 18, 2017

### haruspex

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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