# Homework Help: Finding Gravitational Acceleration from a Slope

1. Jun 12, 2013

### rakeru

1. The problem statement, all variables and given/known data

I have a graph of time squared vs. height. It shows the relationship between the height of a ball and how long it takes for it to fall, squared. I don't have this in the form of a question, but I need to use the slope, which is 0.211 s^2/m, and turn it into a value for acceleration to compare it to the gravitational acceleration. I have no clue how to do this. I did 0.459s/m.. I don't know if that's correct. But I need m/s^2. I have an equation for gravity from a pendulum but I don't think that's what I should use.

2. Relevant equations

I don't have any equations..

3. The attempt at a solution

I got the square root of the seconds squared to get 0.459s/m but I'm not sure if I did that correctly.

2. Jun 12, 2013

### rude man

What is the formula for initial height as a function of time?

3. Jun 12, 2013

### rakeru

Uhm, I don't understand.. So we basically did this in class and we set it up and timed it. We did it multiple times with different heights and then we graphed the data. But it was a quadratic relationship so we were told to use the average time to find time squared and then graph time squared vs height. After that, we got that the slope was 0.211 s^2/m and now I need to use that to find the calculated gravity.. but I don't know how to do that and I wasn't told..

4. Jun 12, 2013

### rude man

5. Jun 12, 2013

### rakeru

...? Does that mean

y=y0+v0t+1/2 at2

That's the only one I know that has initial height..

6. Jun 12, 2013

### rakeru

How could that even be? I don't even know the initial velocity.

7. Jun 12, 2013

### haruspex

Do you mean that you don't know whether it was simply dropped or thrown down, or thrown up? Or do you mean the clock was not started at the moment of release, but after it passed some point? If it was dropped from rest at the moment the clock was started, surely you do know the initial velocity.

8. Jun 12, 2013

### rakeru

oh. my. god. Yeah, I didn't realize that. I still don't get it, though. I have the slope of the line.. and I have different heights, but I only need one answer for the acceleration that counts for everything.

9. Jun 12, 2013

### haruspex

You have 0.211 s^2/m, and you want m/s^2. Square rooting isn't going to work because you'll get square root of metres. Isn't there a much simpler way to turn one of those dimension formulae into the other?

10. Jun 12, 2013

### rude man

That's correct. Initial velocity is zero. You're dropping stuff from rest, aren't you?

11. Jun 12, 2013

### rakeru

I don't know. I really don't know how to change it. I have no clue and I can't even think of anything close because I simply don't know. I can't come up with something out of nothing.

12. Jun 12, 2013

### rakeru

Okay. I think I get it a little after thinking about it for a while. I used that equation and I moved it around to get

a= 2y/t^2

If I were to put the slope there, would it work?? Do I have to somehow make the t^2 on the top?? Please help. is the slope the same thing backwards?? oh my god if this is it i'm going to be so happy.

13. Jun 13, 2013

### haruspex

Yes, it's just a matter of inverting it. x s2/m represents the same relationship as 1/x m/s2. 5 km/h is the same as taking 0.2 h/km.

14. Jun 13, 2013

### rakeru

Wait. What? So is 0.211s^2/m equal to 1m/0.211s^2? :/

So would I have 2m/0.211s^2 ?

15. Jun 13, 2013

### haruspex

Well, they're not equal exactly: you can't substitute one for the other in an equation. But they do represent the same acceleration.
No, how did you get that?

16. Jun 13, 2013

### rakeru

So how would I put that into

a= 2y/t^2 ?

Can't I put it in and do 2*(1m/0.211s^s)?

17. Jun 13, 2013

### rakeru

The height is the independent variable, right?..

18. Jun 13, 2013

### haruspex

Ok, I see. I thought you were magically turning 1m/0.211s^2 into 2m/0.211s^2. Yes, looks ok.

19. Jun 13, 2013

### rakeru

oh my. i almost died. i thought it was right and did my assignment based on it since no one had answered. thank you soooo much!

20. Jun 13, 2013

### rude man

Start with post #10 which reduces to y = at^2/2 since initial conditions are all zero. Change y to h. Now rearrange to t^2 = (2/a)h. Now you have t^2 on the y axis and h on the x axis.

So the slope is 2/a. So now can you determine a from the slope?

EDIT: sorry, I guess you were done before this post.

21. Jun 13, 2013

### rakeru

Omg. So the answer for a=9.48 m/s^2 right?? I'm starting to doubt everything...

22. Jun 13, 2013

### rude man

Why the doubt when you came up with the right answer?

Of course, your data was a bit faulty to begin with because we all know the answer is 9.81 m/s^2 but that's to be expected since you had air resistance and other sources of error to contend with.

Anyway, you're done.