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Initial speed of Stone

  1. Jun 20, 2010 #1
    1. The problem statement, all variables and given/known data
    A stone is dropped off a bridge 50m above the water. Exactly 1s later another stone is thrown down and both stones strike the water together. What must the initial speed of the second stone have been?

    ohh im sorry but i have no idea please help me out here
    thank you soooo much :)
    2. Relevant equations
    conservation law????

    3. The attempt at a solution
  2. jcsd
  3. Jun 20, 2010 #2
    someone plz plz help me :(
  4. Jun 20, 2010 #3
    Try gathering the given data and forming the relevant equations.

    Here, the first stone is 'dropped', and not thrown. Hence, it must be falling down with constant acceleration 'g'. Since you know the height, you can find the time required for the stone to touch the surface of the water.

    Now knowing the time, you can easily find the velocity with which the next stone must be thrown, since the time interval is also given.

    Try this out, and tell us where you get stuck.
  5. Jun 20, 2010 #4
    All you need to use is [tex]s = ut + \frac{1}{2} a{t}^{2}[/tex]

    The only difference between the two cases is with u and t.
    Try it out. It's pretty simple
  6. Jun 21, 2010 #5
    can you please tell me how you got that equation? with that equation it is simple but how did you get it?
  7. Jun 21, 2010 #6


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    Gold Member

    Here is a link showing a derivation, that is non-calculus based:

    The equation is part of a list of equations for "kinematics" equations for constant acceleration. I suggest memorizing each these equations. If you are taking a non-calculus based physics class, your particular coursework/textbook might introduce these equations in a slightly different format from the link above (but they are usually still given to you). Memorize the equations in the format given in your text and coursework (that way, when you use them on homework or a test, there won't be any ambiguity about where you got them).

    If you are taking a calculus based class, the formulas come from the direct result of knowing that (and integrating),

    [tex] \vec {v(t)} = \frac{\partial}{\partial t} \vec {s(t)} \ \ [/tex]

    [tex] \vec {a(t)} = \frac{\partial}{\partial t} \vec {v(t)} \ \ [/tex]
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