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Two objects meeting

  1. Feb 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A stone is dropped into the water form a bridge over top some water. Another stone is thrown vertically upward at 32.0m/s, 2.5 seconds after the first was dropped. Both stones strike the water at the same time. When do the stones meet from the dropping of the first stone?

    2. Relevant equations
    d=vi+1/2at^2
    SUVAT equations
    3. The attempt at a solution
    d=1/2(9.8)(t)^2
    =4.9t^2
    This is for stone 1. I don't know how to setup the other equation since the distance traveled is not the same as stone 1.
     
  2. jcsd
  3. Feb 15, 2015 #2

    Nathanael

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    The second stone is thrown upwards and after the first, and they still end up meeting? Doesn't that seem a bit strange?

    The same way you did for stone one, but you use a different variable for time (say, capital T). There is a simple relationship between t (the time when the first stone starts it motion) and T (the time when the second stone starts it's motion). Do you know what it is?
     
  4. Feb 16, 2015 #3

    Suraj M

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    OP, d in the equation$$ d = \frac{1}{2}at² $$
    is not distance travelled by the particle,its the displacement, which in this case, is same for both the stones, so you can equate them, think of what Nathanael said, find the time for of flight for the 2nd stone,equate and solve for t.
     
    Last edited: Feb 16, 2015
  5. Feb 16, 2015 #4

    HallsofIvy

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    Are you sure you have stated the problem correctly? If so then, as both Nathanael and Suraj N suggest, the "first time they meet" must be when they hit the water.

    In any case, you can as you say, use the formula for motion under constant acceleration, (1/2)a(t- t0)^2+ u(t- t0)+ d where a is the constant acceleration, u is the initial speed, d is the initial height, and t0 is the time the motion starts. In order to use that, of course, you need to set up a "coordinate system". The standard thing to do is to take + upward, height= 0 at the surface of the water, and t0= 0 when the first stone is thrown. "A stone is dropped into the water form a bridge over top some water". So -9.8 m/s^2, u= 0, and we are not told how high the bridge is so leave that as "d0". d= -9.8t^2+ d0.

    " Another stone is thrown vertically upward at 32.0m/s, 2.5 seconds after the first was dropped." Now, a= -9.8, u= 32, t0= 2.5, and d is still d0. d= -9.8(t- 2.5)^2+ 32t+ d0

    "Both stones strike the water at the same time." So at some time t, both of those are equal to "d= d0" so are equal to each other: -98t^2+ d0= -9.8(t- 2.5)^2+ 32t+ d0.
    That reduces to a linear equation for t. The two stones will have the same height for only one value of t, the time they hit the water together.[/I]
     
  6. Feb 16, 2015 #5
    Would it make a difference if the second stone is thrown from the bottom? Like if one is dropped from the top, and the other is thrown upwards from the bottom.

    Yes, I see what you're saying. I don't think it makes sense in the question, but that is what it says. I think the question is trying to say the second stone is thrown from a different position (bottom of the bridge).
     
  7. Feb 16, 2015 #6

    lightgrav

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    There are 2 equations ... so there can only be 2 free variables to solve for.
    You're told the initial velocity for both stones (v1=0), but neither starting location, and neither ending location.
    Because ending locations are the same, there can only be 1 "free" starting location
    this means that the distance below the bridge that the 2nd stone must begin
    is not free to vary ... it can be calculated (as part "b" for this scenario) after "t" is obtained.
     
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