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Two stones having a meeting

  1. Aug 31, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose you throw a stone straight up with an initial velocity of 20.5 m/s and, 4.0 s later you throw a second stone straight up with the same initial velocity. The first stone going down will meet the second stone going up. At what height above the point of release do the two stones meet?


    2. Relevant equations

    xf = xi + vit + (at2)/(2)

    3. The attempt at a solution

    So far I have setup an equation for each stone involving 't' and set them equal to find the time the stone meets in the air, and then attempted to solve for the height of either stone at time 't'. My answer in consistently 1.84 meters. I keep getting told this is the wrong answer.

    So far I have this
    vt-gt^2 = v(t-4) - g(t-4)^2
    simplified to:
    t = (v/g) + 2
    Plug in numbers to get:
    t = 4.0918;
    plug 't' into the position function for stone 1:
    height = 20.5(4.0918)-9.8(4.0918^2) = 1.84 meters.

    I'm not sure whats going on, I have also tried using (t+4) for tone #2, but the problem still turns out to be wrong.
     
  2. jcsd
  3. Aug 31, 2014 #2

    SammyS

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    (Fixed some typos in Edit.)


    Hello Matthematic. Welcome to PF !

    Thanks for using the template correctly.

    Why would you use t + 4 for anything?


    Show some steps in going from

    vt - (g/2)t2 = v(t-4) - (g/2)(t-4)2

    to

    t = (v/g) + 2​

    To quickly check your answer, plug the value of t that you get into each side of

    vt - (g/2)t2 = v(t-4) - (g/2)(t-4)2

    to verify that the result makes sense.
     
    Last edited: Aug 31, 2014
  4. Aug 31, 2014 #3
    Math fun

    I thought maybe my mistake was that I should calculate as t=0 for first stone, and t=4 for the second stone (t and t+4). I thought wrong.

    vt+(-[itex]\frac{g}{2}[/itex])t2=v(t-4)+(-g)(t-4)2
    vt - [itex]\frac{g}{2}[/itex]t2=vt - 4v - [itex]\frac{g}{2}[/itex]t2 + 4gt - 8g //expand and distribute.
    0 = -4v + 4gt -8g // cancel out terms that exists on both sides.
    4v + 8g = 4gt // Move terms over to the other size to isolate the 't'
    [itex]\frac{v}{g}[/itex] + 2 = t // divide by 4g.
    Put in 20.5 s 'v' and 9.8 as 'g'
    t = 4.0918
    Now use 't' in the position function to determine what height the stones are at at the time that they are the same height.
    s(t) = 20.5(4.0918) - [itex]\frac{9.8}{2}[/itex](4.09182) = 1.8413 meters. (wrong answer)

    At the time of my writing this, I had a thought. Is that function above the velocity function or the position function? I thought it was the position function. Perhaps my error is that I should integrate the function and then plug in t? I'm not to sure, I only have one more try or I get marked wrong for the problem.
     
  5. Aug 31, 2014 #4

    SammyS

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    Everything looks fine.

    Do you get the same answer plugging into
    v(t-4) - (g/2)(t-4)2 ?​
     
  6. Aug 31, 2014 #5
    Alternative formulae

    Yes.

    Update:

    Using

    vf = vi +at2

    and

    d = ((vi + vf)/2) *t

    I have calculated displacement again getting the same answer.

    Edit: I think its time to move on. When I get the solution from the professor or a student who got it right, ill post it here for any future googlers who find this tragic story.
     
    Last edited: Aug 31, 2014
  7. Aug 31, 2014 #6

    BvU

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    Could it be as simple as you having to use g = 9.81 m/s2 and then getting someting like 1.80 m ?
    In which case you've done everything right, but different from what's in the answer book.... c.q. the poor computer.
     
  8. Aug 31, 2014 #7

    SammyS

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    How do you know it's the wrong answer.

    If this is graded by some website homework service, or the like, it may be as simple as a problem of an incorrect number of significant digits.
     
  9. Sep 1, 2014 #8
    My professor is doing me a favor because I turned it in early, he said I can still try one more time to get it right. Ill try the more accurate gravity constant. But ill make sure to figure out what went wrong and post it.
     
  10. Sep 3, 2014 #9
    I made a guess based on the sig-figs of the other numbers and used 1.80 (instead of 1.84) and I got it right. This is gonna be a long semester if more stuff like this happens.
     
  11. Sep 3, 2014 #10

    BvU

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    No big deal: you did the right things, only used 9.8 instead of 9.81. Semesters look a lot shorter once you gather some experience with them :smile:
     
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