# Homework Help: Sky Diving Analysis

1. Nov 10, 2009

### rahul129

1. The problem statement, all variables and given/known data
Members of a sky diving club were given the following data to use in planning their jumps. In the table attached, d represents the distance fallen from the rest by a sky diver in a "free-fall stable spread position" verses the time of the fall.Analyze bot qualitatively and quantitatively the motion of a sky diver using the data provided. Draw graphs for position,velocity, and acceleration vs. time.

2. Relevant equations
None that i know to use.

3. The attempt at a solution
I am kind of a noob at physics so this may be wrong. Quantitatively, the data is increasing, but in different rates. The position vs. time graph would therefore look like an exponential function. Qualitatively, the equation for the data (through a QUAD regression on TI-83) gave me 2.36x2+108.44x-186.32. I do not know how to figure out the velocity or acceleration vs. time graph.

Honestly I have no idea if this is right. Please someone help! Thanks in advance!

#### Attached Files:

• ###### skydive table.doc
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Last edited: Nov 10, 2009
2. Nov 10, 2009

### ideasrule

If you thought the position vs. time graph looked like an exponential function (which it is actually is), why did you do a quad regression?

As for v vs. t and a vs. t graphs, how do you calculate velocity from position? Do this to generate a list of velocities. How do you calculate acceleration from velocity? Do this to generate a list of accelerations.

3. Nov 10, 2009

### rahul129

I chose to do a quad regression because it fits the data better. I know this is weird but i dont know why that worked out. Anyways, how would i generate the velocity from position? I was thinking of finding the derivative at each point. I'm not sure if thats correct though.

4. Nov 10, 2009

### ideasrule

Finding the derivative at each point is the most accurate way to do it, but it's going to be quite tedious. A simpler way is to calculate the velocity in between, say, 1 s and 2 s as (d2-d1)/(2s - 1 s) and assume that this is equal to the velocity at t=1.5 s. This is only strictly true if acceleration is constant, which it isn't in this case, but the result will be fairly accurate.

5. Nov 10, 2009

### rahul129

Thanks! I have a few more questions though. Can't I just take the derivative of the whole function and plug the values of t back in? How would I analyze this quatitatively and qualitatively? Also I was playing around with the graph in excel and putting the function in a cubic or 4th degree regression yielded a higher R squared value should I use these or the quadratic?