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Kinematics - When do they meet?

  • Thread starter Sean1218
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  • #1
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1. Homework Statement

A stone is dropped off a cliff of height h. At the same time, a second stone is throne straight upward from the base of a cliff with an initial velocity v. Assuming that the second rock is thrown hard enough at what time t will the two stones meet.

2. Homework Equations

Uniform acceleration equations

3. The Attempt at a Solution

I tried a few different things, and nothing worked. I don't have anything written down right now though.

I just wrote Δd = v1Δt + 1/2gΔt2, for both stones, substituting Δd for x in one of the equations, and h - x in the other.

So, I just had x = 1/2gΔt2 and h - x = v1Δt + 1/2gΔt2

I tried combining them, but I got to a point where I didn't think I could do anything else with it, and it just felt wrong.

Am I thinking about it wrong?
 
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Answers and Replies

  • #2
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if you told me the numbers i could solve it for you :p
 
  • #3
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if you told me the numbers i could solve it for you :p
There are no numbers, could you explain how to solve it?
 
  • #4
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let me get some paper i gtg eat so ill tell you when i find out
 
  • #5
tiny-tim
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Hi Sean! :smile:
I just wrote Δd = v1Δt + gΔt2, for both stones, substituting Δd for x in one of the equations, and h - x in the other.

So, I just had x = gΔt2 and h - x = v1Δt + gΔt2
Yes, you have the correct basic idea, but

i] it's 1/2 g∆t2, isn't it?

ii] shouldn't one of those gs be minus?

(btw, you'll notice something unexpectedly disappears … why do you think that is?)

(and jrab, on this forum you mustn't give answers, you must only help)
 
  • #6
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ah haha, forgot the 1/2. That doesn't really change my result much though.

I still end up with 0 = gΔt2 + v1Δt - h, and I don't see how I can use the quadratic formula without numbers. Nothing disappeared or cancelled out from what I can tell, since moving the - 1/2gΔt2 to the other side just makes it positive. And why would one of the g's be negative? I'm using v1 for both, so the equation is just + g.
 
  • #7
tiny-tim
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But isn't your v1 up, and your g down?
 
  • #8
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Not really sure what you mean. v1 is going up, and g is down, but I can't just include negatives can I? g = -9.8, if I made g negative, at the end I'd just get a positive acceleration, but g isn't positive. g is the same for both equations, v1 = 0 for the first one, and it's positive for the second stone, but I don't see how I can declare that it'll always be positive without plugging in numbers, if you know what I mean.
 
  • #9
tiny-tim
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Oh i see, you're using g = minus 9.8 (we don't usually do that) …

ok, but then if your x is always positive, doesn't the first g need to have a minus?

Anyway, here's another approach … what is the relative acceleration of the two stones? :wink:
 
  • #10
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The first equation is the same as the second equation though. v1 is 0 so the first term is 0, but g is still g (-9.8 if you plug it in) as far as I know.

and I wouldn't know where to start with relative acceleration
 
  • #11
tiny-tim
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… I wouldn't know where to start with relative acceleration
Well, what is the acceleration of the first stone, and what is the acceleration of the second stone?
 
  • #12
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it's the same for both, accel due to gravity.
 
  • #13
tiny-tim
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So the relative acceleration is … ? :smile:
 
  • #14
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relative to each other, 2g I guess
 
  • #15
tiny-tim
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Nooo! it's zero, isn't it?

And if the relative acceleration is zero, then the relationship between relative distance relative time and relative speed is … ? :wink:

I'm off to bed now :zzz: …
 
  • #16
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Can anyone else help with the original method I was using (after adding in the 1/2 that is)?
 
  • #17
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Anyone?
 
  • #18
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an easier way would be relative motion as tiny-tim suggested.
And accordingly you can plug in the equations of motion considering the relative acceleration as 0.
rel(d) = rel(speed) * time.

easy as that.
 
  • #19
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So, I just had x = 1/2gΔt2 and h - x = v1Δt + 1/2gΔt2
Using g=+9.8m/s^2
The second one in there would have a= -g.

Now add them up, eliminating few stuff you can get your answer, your way.
 
  • #20
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Using g=+9.8m/s^2
The second one in there would have a= -g.

Now add them up, eliminating few stuff you can get your answer, your way.
Alright, thanks! So, why would one of them be -g instead of g? Accel due to gravity is always negative, so I don't really see why.
 
  • #21
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we dont usually use g to be negative, the direction decides what g is. Only the thing is g acts downwards, and if you choose up to be positive g becomes negative and if you choose down to be positive g becomes positive.

So get it?
 
  • #22
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Not really, I was working through it assuming up is positive, and down is negative, for both equations. Was there something I did that indicated otherwise?
 
  • #23
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then in your first equation x would be negative....
 

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