I'm most of the way through the physics homework I have due next week. The only problem I'm hung up on is a proof of an equation for finding the acceleration of gravity from an experiment in which a ball is thrown upward in a tube that's been evacuated of air. It passes two marks, separated by a known distance, on the way up and down, and the times it takes to pass the lower mark twice and the upper mark twice are recorded. I believe that the information to find the acceleration of gravity is in there, but I'm having a hell of a time figuring out how to express it without needing to know the initial velocity, or maybe the height at which the ball stops moving before traveling down again.

Differential equations, though, are another level of hard. I'm probably going to spend tomorrow afternoon in the library again, where I can borrow the text book and read and re-read the chapters and example problems. I should get that book next week. I guess if I had to choose one book not to have a copy of, it would be physics - everything so far is a review of high school physics, although the class does move faster, and I can photocopy the homework assignments from a classmate's book until I get mine.

The whole unemployment thing continues. It's as confusing as differential equations, but the money at stake is in a short-term rather than long-term time frame.

## 6 comments:

You might be able to do at as a system of equations of the form of

y= vt + at^2/2 + c

You have y and t at each of 4 points, so you need to solve for v and a--c drops out. You can assume that v is the same going down at each of the two heights as it was going up (conservation of energy), which helps two variables drop out. That should give you all the information you need.

Hmm. Did they give you an answer? I get (given that the lower mark has a height of 0 and the upper mark a height of d and that t1-4 are the consecutive recorded times the ball passes each mark) a = 4d/(t2^2-t1^2+t3^2-t4^2).

Man, it's been a while.

I solved it this morning, over coffee.

I figured out I actually did know velocities, and that was the inspiration that helped me crack open the problem - I can calculate, at least symbolically, the velocity when the ball passes each mark on the downward journey because I know it stops at the highest point. So I can also integrate again and get the displacement.

I ended up never using velocity per se. I also know that the distance between the two lines, 'H', is the definite integral of gt from the time at which it passes the upper line to the time at which it passes the lower line. Integrating, substituting manipulations of the differences in time I'm given, and then solving for g gave me the answer given in the handout.

g = 8H/(tl^2 - tu^2)

tl and tu are the times between the first and second passage of the lower mark and upper mark respectively.

Trying to figure out an average velocity was a little bit of a false lead for me - I never needed to know it. I also didn't need to use the velocity at the upper and lower marks once I figured out that they were both a function of gravity and time and it was enough to know the times of the ball's passage.

Now that I've spent so much energy on this and you've contributed to, I'm going to hunt down my MS Office install disc and see if the equation editor has the right stuff for me to publish calculus functions legibly. Watch this space!

Also, I think if I set up the problem with your labeling at each point, I'll get your answer and also be able to turn it into their answer by substituting manipulations of their labeling for your (clearer, and I used something similar myself) labels.

Wait, you integrated? You're crazy.

I liked the cleanliness of the integral. It sets up as a single term on each side, and there are never more than two. Also, I've been doing nothing but integral Calculus for the last few months, so it's the first tool that springs to mind.

Kind of like putting up drywall with an 18V DeWalt, but whatever.

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