Finding Gravitational Acceleration from a Slope

[rakeru]

Homework Statement

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.

Homework Equations

I don't have any equations..

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.

Pleaaasse Help. Please.

 

[rude man]

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

 

[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..

 

[rude man]

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..

Answer my post.

 

[rakeru]

Answer my post.

...? Does that mean

y=y₀+v₀t+1/2 at²

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

 

[rakeru]

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

 

[haruspex]

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

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.

 

[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.

 

[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?

 

[rude man]

...? Does that mean

y=y₀+v₀t+1/2 at²

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

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

 

[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.

 

[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.

 

[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.

 

[rakeru]

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

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

 

[haruspex]

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

Well, they're not equal exactly: you can't substitute one for the other in an equation. But they do represent the same acceleration.

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

No, how did you get that?

 

[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)?

 

[rakeru]

The height is the independent variable, right?..

 

[haruspex]

So how would I put that into

a= 2y/t^2 ?

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

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

 

[rakeru]

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

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!

 

[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.

 

[rakeru]

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

 

[rude man]

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

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.