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 Physics: Kinematics
 
Dieyou2000
Posted: October 20, 2009 10:31 pmTop
   
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 I'm pretty sure it's called kinematics anyway. Well yeah so basically I'm in physics, which is "taught like a college class". Apparently this means "never lecture or instruct the class" to my teacher and I've been able to learn most of the stuff on my own but there is this one thing I can't get, and he's no help so I've come here.

A sample problem would be something like:

A cannon shoots a cannonball at a speed of 50.0 m/s at an angle of 25 degrees from the ground.

A) Find the vertical and horizontal components.

I can find the components that isn't hard at all.

cool.gif How long does the cannonball stay in the air?

Now I know I only use the vertical factors. So I would use -9.81 m/s(squared) for acceleration, time is the variable I'm trying to find, and I'm pretty sure initial velocity is just the 50.0 m/s. I have no idea about final velocity or whether I need distance.

C) How far does the cannonball go horizontally?

Again I know to separate it into only horizontal factors, but past that I have no clue what to do.

D) What is the maximum height the cannonball reaches?

*starts slamming head against desk*

The forumulas we're using are:

vf=vi+at
x=vi(t)+1/2a(t(squared))
vf(squared)=vi(squared)+2ax
x=1/2(vi+vf)t

vf=final velocity
vi=initial velocity
x=displacement/distance
t=time
a=acceleration

I'm pretty sure you're supposed to use the initial velocity combined with the angle too? So like it would be 50.0(cos 25) or something like that?

Other major thing I don't get is how to use initial and final velocities work. I get that if it's the maximum height you use 0 for final velocity, but other than that I'm clueless.

Any help appreciated.

P.S. - Fuck my teacher.
 
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Samurai-JM
Posted: October 20, 2009 10:33 pmTop
   
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 physics is easy
 
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W I N N I N G
Kyle
Posted: October 20, 2009 10:33 pmTop
   
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 wat
 
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Planolocal
Posted: October 20, 2009 10:35 pmTop
   
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 s0z br0z i dont kn0wz
 
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"If at first you don’t succeed, call it version 1.0"
Dieyou2000
Posted: October 20, 2009 10:39 pmTop
   
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QUOTE   Samurai-JM @ October 20, 2009 05:33 pm)

physics is easy

Then tell me how to do it. Honestly not in the best mood right now as I probably just failed a physics test because my teacher is a prick.
 
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Chappell44
Posted: October 20, 2009 10:48 pmTop
   


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 confused =[
 
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Fullmetal792
Posted: October 20, 2009 11:17 pmTop
   
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 Okay, I was pretty bad with projectile motion, which is what this is, but I'll give it a shot...

You sound like you got A, B asks for time, so I think we would have to use a formula like t=(vf-vi)/a where...

vi = horizontal component, I believe
a = -9.81 m/s^2
t = ?
vf = 0 m/s

vf = 0 because you're looking for the time it takes for the thing to hit the ground (unless it bounces, which I doubt a cannonball would do) and then solve for time.

3) For this, you need the time in the last answer to find distance. It should be your horizontal component times Time. If it was from a top of a building, I might be able to help more, but then you'd see a wall of text and I'd probably scare my dogs with my cursing trying to do this problem with no scratch paper. neko2.gif

4) Maximum height is using the vertical component in the formula used to find the answer to 1. You then use that answer and plug time into (1/2)at^2. "a" should be 9.81m/s

I think thats how all of these are done... You can do the work for it and see if it all works out.

Don't worry about your physics stuff this early on in the year, I promise you it'll get better. Hopefully by the end of the year you'll be on electricity and that is SOOOOOO easy. I could do it in my sleep. Advice, get a graphing calculator if you don't have one, there are some very VERY useful functions on some of them (TI-83s/84s cost a lot, but still have some cool functions I could show you if you have one of them), and the next big thing you probably will have to worry about is Buoyancy or friction. Turning stuff is pretty annoying too, but you'll live, I hope.
 
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Thanks Sam for an awesome job and an awesome wait!
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Dieyou2000
Posted: October 21, 2009 03:15 amTop
   
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 Thanks FullMetal and Robbie I understand what I'm supposed to do now. Just need some practice and I'll be good.

Thanks a ton guys. happy.gif wub.gif
 
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RobbieThe1st
Posted: October 21, 2009 03:23 amTop
   
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QUOTE   Fullmetal792 @ October 20, 2009 11:17 pm)

Okay, I was pretty bad with projectile motion, which is what this is, but I'll give it a shot...

You sound like you got A, B asks for time, so I think we would have to use a formula like t=(vf-vi)/a where...

vi = horizontal component, I believe
a = -9.81 m/s^2
t = ?
vf = 0 m/s

vf = 0 because you're looking for the time it takes for the thing to hit the ground (unless it bounces, which I doubt a cannonball would do) and then solve for time.

3) For this, you need the time in the last answer to find distance. It should be your horizontal component times Time. If it was from a top of a building, I might be able to help more, but then you'd see a wall of text and I'd probably scare my dogs with my cursing trying to do this problem with no scratch paper. neko2.gif

4) Maximum height is using the vertical component in the formula used to find the answer to 1. You then use that answer and plug time into (1/2)at^2. "a" should be 9.81m/s

I think thats how all of these are done... You can do the work for it and see if it all works out.

Don't worry about your physics stuff this early on in the year, I promise you it'll get better. Hopefully by the end of the year you'll be on electricity and that is SOOOOOO easy. I could do it in my sleep. Advice, get a graphing calculator if you don't have one, there are some very VERY useful functions on some of them (TI-83s/84s cost a lot, but still have some cool functions I could show you if you have one of them), and the next big thing you probably will have to worry about is Buoyancy or friction. Turning stuff is pretty annoying too, but you'll live, I hope.

I see a major mistake: When velocity is 0 is not when it hits the ground, but at the peak height of the projectile's flight. From there, you simply need to know that the time to the peak is the same as the time from the peak to the ground.

Calculator wise, I suggest a TI81, 82 or 85. They look boxy and old, and should be between $10 and $20 on ebay. They aren't the fastest, but are great calculators for all sorts of things, and graph fast enough anyway.
They are also near-indestructable. I have a TI81 that I use in the machine shop at school, I have dropped it quite a few times, its always getting coolant, oil and water splattered on it.. it still works great. Also, once I found out how to adjust the screen contrast(second function + UP/DOWN key), I haven't had any trouble with batteries.

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I'm helping Dieyou2000 in IRC right now, only posting this as a followup.

for A), we use trignometry:
sin(25)*50.0 = Y component = 21.130913087
cos(25)*50.0 = x component = 43.3153893518

For cool.gif, we take formula 1, and fill it in. we use Vf = 0 to get the point at the peak:
vf=vi+at
0 =21.130913087-9.8T
so
21.130913087 = 9.8T
T = 2.15621562112

Now, for the time to the ground, we double that:
TTG = 2.15621562112 * 2 = 4.31243124224

For C), we take formula 4 and simplify it
x=1/2(vi+vf)T
simplifies to
x=vT if velocity is constant, so
x = 4.31243124224 * 43.3153893518 = 186.79463831


For D), lets use:
x=vi(t)+1/2at^2
so
x = 21.130913087*2.15621562112+.5*(-9.8)2.15621562112^2
x = 45.5628048867 + .5*-9.8*4.64926580476
x = 45.5628048867 -22.7814024433
x = 22.7814024434

If we used:
x=1/2(vi+vf)t
x = .5*(21.130913087+0)*2.15621562112
22.7814024434
We get the same answer.




 
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Fullmetal792
Posted: October 21, 2009 03:41 amTop
   
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 Thanks for pointing that out Robbie, like I said, I was bad with projectile motion last year... I also got sick for the week we went over it, came back and had a test over it. I got a 40 from guess work. Then Ike hit and I just stopped caring about physics. Ended up dropping the hard class and getting the better teacher.

I didn't think I did 2 right originally since I went back and checked it in my calculator and saw my answer was off. I was better with sound levels, electricity and current, and waves anyway, lol.
 
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Thanks Sam for an awesome job and an awesome wait!
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But you sir, are a legend
kajju
Posted: June 20, 2010 03:33 amTop
   
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 i done intermediates exams but never met a problem like this wtf are thesE???@!?$!@?
 
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Dieyou2000
Posted: June 20, 2010 04:19 amTop
   
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 Don't bump old, useless topics.
 
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