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Old 02-14-2006, 11:01 PM   #1
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Physics help... I'm stuck

I have this problem for an online hw set. Not due till Friday, but we won't reach this material in class and I don't like to wait. I can't find any help at all on this.

Problem:
A stone is thrown vertically upward at a speed of 37.30 m/s at time t=0. A second stone is thrown upward with the same speed 2.260 s later.



I need:

a) At what time are the two stones at the same height?

b) At what height do the two stones pass each other?

c) What is the upward speed of the second stone as they pass each other?

d) What is the downward speed of the first stone as they pass each other?
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Old 02-14-2006, 11:53 PM   #2
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Re: Physics help... I'm stuck

meh, thats tricky. the easiest way i can think of doing it is first calculate when the first stone is at max height. i dont have the formulas with me but when the stone is at 0m/s and starting to head downward thats where id start the problem off. calculate stone 2's height and speed and then start the problem there.
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Old 02-14-2006, 11:57 PM   #3
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Re: Physics help... I'm stuck

the maxheight and speed at maxheight is going to be the same for both stones.

I suppose the time at maxheight would be different though... let me go back to work.
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Old 02-15-2006, 12:23 AM   #4
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Re: Physics help... I'm stuck

I'm bored and have no life, so let's see what I can do.

Question a, at what time does h1 = h2?

Ok, we're given:

t1 = 0
t2 = 2.260


We know that h = (1/2)a(t^2) + vt, where a is acceleration (in this case g, gravity), v is the initial velocity, and t is the time of travel.

Note that the actual elapsed time of travel at some time t, will be t - t1 (or just t because t1 = 0) for stone 1 and t - t2 (or t - 2.260) for stone 2.

We need to find the time at which h1 = h2, so...

(1/2) * g * (t)^2 + v * t = (1/2) * g * (t - 2.260)^2 + v * (t - 2.260)

.5g(t^2) + vt = .5g(t^2 - 4.520t + 5.17076) + vt - 2.260v

.5g(t^2) + vt = .5g(t^2) - 2.260gt + 2.58538g + vt - 2.260v

2.260gt = 2.58538g - 2.260v

t = (2.58538g)/(2.260g) - (2.260v)/(2.260g)

t = (2.58538 / 2.260) - (v / g)

t = (2.58538 / 2.260) - (37.30 / -9.8)

t = 4.95 seconds

For question b, we can simply plug in this elapsed time at which the two pass into h = (1/2)a(t^2) + vt.

So we get:

h = (1/2) * g * (4.95 ^ 2) + v * 4.95

This comes out to 64.57275 m.

You could also use:

h = (1/2) * g * ((4.95 - 2.260) ^ 2) + v * (4.95 - 2.260)

This number may be just slightly off from the first result simply because of rounding issues (especially the use of 9.8 m/s^2 as the value of g).

The math is a little messy b/c of the squaring of the difference in time, but other than that is pretty straight-forward. You may need to use a more specific value for g, different teachers expect different things.

Problems c and d shouldn't be too difficult from here, now that you know the time elapsed when the two pass. Just remember that stone 2 has actually only been traveling (4.95 - 2.260) seconds when they pass and not the full 4.95.

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Old 02-15-2006, 12:27 AM   #5
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Re: Physics help... I'm stuck

dammit, i felt all smart, haha. I just got the time at the same height, and it is 4.95 seconds.

Thanks for the help man.
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Old 02-15-2006, 12:32 AM   #6
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Re: Physics help... I'm stuck

I went the long way about it though.

I drew a(n approximate) graph of thw two stones, x vs t. I plotted TimeMaxH1 and TimeMaxH2, because I solved for those, getting 3.8s for 1 and 6.07s for 2. I subtracted 3.8s from 6.07s, getting 2.27s, which i halved and added to 3.8s, giving me the time between the two maxheights, and as the graphs are the same, just at different times, the halfway point between the two times is where the position intersects.
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Old 02-15-2006, 12:35 AM   #7
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Re: Physics help... I'm stuck

my teacher uses -10 m/s^2 as gravity and will write an answer like 8327120 m in the classroom, then gives us online hw that requires us to use -9.8m/s^2, and have no more than 2 sig figs. he's gay.
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Old 02-15-2006, 12:47 AM   #8
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Re: Physics help... I'm stuck

No problem.
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Old 02-15-2006, 12:54 AM   #9
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Re: Physics help... I'm stuck

C) 11.036 m/s

V = Vo + at

V=37.3 m/s
Vo=0m/s
t=2.68s (4.95s - 2.26s)


D)11.11 m/s

I suppose that C and D should be equal, but there's that whole rounding thing.
D uses the same formula as C, except 4.94s for the t instead of 2.68s. Also, D is apparently a scalar and not a vector, because the question specifies that the movement is downward, so no need for a negative. :dunno:
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Old 02-15-2006, 01:05 AM   #10
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Re: Physics help... I'm stuck

Wow... my head hurts
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Old 02-15-2006, 01:35 AM   #11
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Re: Physics help... I'm stuck

Hey guys..sorry to interrupt this thread...but I don't know how to start a new thread.

I have a 2003 mustang GT with the interior upgrade package....which includes an aluminum ring around my shifter. My headlight knob broke..actually came off....quite EASILy..and ACCIDENTALLY for me

NOW..I want to get one of those metal knobs..but I don't know what finish to get...silver anodized...chrome polished....machine finish...I want it to match my shifter ring...ALSO..the little clip to install the original knob broke...do the "new" knobs come with a replacement clip??..Any info would be appreciated..as I hate using needlenose to turn on my headlights..lol...

Thanks
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Old 02-15-2006, 01:56 AM   #12
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Re: Physics help... I'm stuck

Hmm... how exactly did you calculate C & D? I'm getting something slightly different....

First, I found the time it would take for a stone to reach it's maximum height (which will be where it's velocity reaches zero). This must be done because of the fact that stone 1 has reached its peak and began to fall back. This falling must be viewed separately from the travel upwards when using V= v0 + at. On the move upwards the initial velocity will simply be the initial velocity given in the problem. On the fall, however, the stone will start with a velocity of zero and end up with negative velocity.

So plugging in what we know, we get:

0 = 37.30 + (-10)t
10t = 37.30
t = 3.73 seconds

So it takes 3.73 seconds for each stone to reach its maximum height. Remember, this is important because the initial velocity will be different depending on whether the stone in question is going up or down. If the stone has been travelling for more than 3.73 seconds (i.e. stone 1), it's coming down. If it has been travelling less (i.e. stone 2), it's still going up.

So for stone 2 (Question C) using V= v0 + at once again, the initial velocity will simply be the velocity in which it was propelled up.

So we have:

V = 37.30 + (-10)(4.95 - 2.260)
V = 37.30 - 26.90
V = 10.4 m/s

Now, stone 1 has already reached its peak and began to fall back. This means its initial velocity is going to be zero. It's velocity has reached zero and it will now begin accelerating back down to Earth. So once again, we use V= v0 + at, this time with an intial velocity of zero. Also, you have to remember that for the first 3.73 seconds this stone was going up. To get the time it has been accelerating back down, you have to subtract these 3.73 seconds from its total time of travel.

So we get:

V = 0 + (-10)(4.95 - 3.73)
V = (-10)(1.22)
V = -12.2 m/s

Hope this helps. Maybe I should give up this whole programming thing and go teach physics somewhere....
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Old 02-15-2006, 02:03 AM   #13
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Re: Physics help... I'm stuck

Quote:
Originally Posted by MelissaGT
Hey guys..sorry to interrupt this thread...but I don't know how to start a new thread.



Thanks

go to the forum index, click on whatever forum you want to post in... then click "new Thread"

Type your message... and done.


Remember, just ask if you have any questions
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Old 02-15-2006, 08:23 AM   #14
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Re: Physics help... I'm stuck

that was a good explaination, gotta love physics
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Old 02-15-2006, 09:11 AM   #15
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Re: Physics help... I'm stuck

Quote:
Originally Posted by WhiteStang99
C) 11.036 m/s

V = Vo + at

V=37.3 m/s
Vo=0m/s
t=2.68s (4.95s - 2.26s)


D)11.11 m/s

I suppose that C and D should be equal, but there's that whole rounding thing.
D uses the same formula as C, except 4.94s for the t instead of 2.68s. Also, D is apparently a scalar and not a vector, because the question specifies that the movement is downward, so no need for a negative. :dunno:
they don't have to be equal... they are two independant events, you're just tying them together by finding the differences. Also, I wish I had seen this problem last night .... I would've loved to tackle it. And yes, you do need the negative vector... or at least we did, when I was a physics major.
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Old 02-15-2006, 12:30 PM   #16
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Re: Physics help... I'm stuck

No, you don't need a negative Jimmy, the question specified that the movement was down, so adding the negative would make the movement in the positive y axis, which would be going up. This is what I was doing last night when you kept IMing me about motorcycles, haha.

I thought I had explained how I got my answers in that last post, but it doesn't really matter, I got it all right w/ the answers I posted.


Thanks a lot bbunt. I figured out the time right before I checked the thread again, and I probably could have figured out the rest with that information, but your help was greatly appreciated.
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Old 02-16-2006, 12:53 AM   #17
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Re: Physics help... I'm stuck

i'm asian and all i see is 'blah blah numbers omgwtf does it mean'
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Old 02-16-2006, 04:15 AM   #18
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Re: Physics help... I'm stuck

my head hurt now
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Old 02-16-2006, 08:26 AM   #19
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Re: Physics help... I'm stuck

i never too physics...and that just made my head hurt...
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Old 02-16-2006, 11:30 AM   #20
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Re: Physics help... I'm stuck

Quote:
Originally Posted by WhiteStang99
No, you don't need a negative Jimmy, the question specified that the movement was down, so adding the negative would make the movement in the positive y axis, which would be going up. This is what I was doing last night when you kept IMing me about motorcycles, haha.

I thought I had explained how I got my answers in that last post, but it doesn't really matter, I got it all right w/ the answers I posted.


Thanks a lot bbunt. I figured out the time right before I checked the thread again, and I probably could have figured out the rest with that information, but your help was greatly appreciated.
we had to use the negative because our answers had to "stand on their own, without regard to the question"... but our teacher was Italian... what do you expect?
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Old 02-16-2006, 11:45 AM   #21
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Re: Physics help... I'm stuck

Ours is Indian, and no **** the first day of class came in and says:
Quote:
I am Doctor Sadanandandan Vhallabanennini, but you call me Doctor Rao, because you cannot pronounce my name.
And that's his real name, i'm looking at the syllabus to make sure it's spelled right, haha.
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Old 02-16-2006, 04:11 PM   #22
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Re: Physics help... I'm stuck

Quote:
Originally Posted by WhiteStang99
Ours is Indian, and no **** the first day of class came in and says:


And that's his real name, i'm looking at the syllabus to make sure it's spelled right, haha.
impressive... our's was Mario (go figure... Italian) Affagatiato.... I'm guessing, but that's quite close.... we called him Mario. The other guy was Dr. Feller.. we called him Doc, not because Feller is so hard, but because he's a really cool guy from New York.
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