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Stone dropping - yet another free fall acceleration problem

  1. Sep 11, 2003 #1
    I'm sorry, I am just having difficulty with being sure of myself when it comes to these problems. This one states:

    A stone is dropped into a river from a bridge 43.9m above the water. Another stone is thrown vertically down 1.00 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?

    Well, I did this:

    Y-Yo = 43.9m
    a = -9.8 m/s^2

    Vo1 = 0 m/s because it was just dropped

    Y-Yo = Vot + 0.5at^2
    43.9 = 0 + 0.5(9.8)t^2 = 4.9t^2 --> t1 = 2.99s

    Since they both reached the water at the same time, t1 = 2.99s = t2.
    Since stone 2 is thrown 1.00 s after stone 1, I decided to do: 2.99 + 1.00 = 3.99s to use for t in Y-Yo = Vot + 0.5at^2 for stone 2

    So I got:

    43.99 = Vo(3.99) + 0.5(-9.8)(3.99)^2 --> Vo = 30.5535 m/s

    For some reason, I believe my method to be wrong because doesn't the t in Y-Yo = Vot + 0.5at^2 mean the total time it took to reach the water? And since they both reached the water at the same time, how can the t I used the second time be 3.99s? If I used 2.99s for t to get the Vo for stone 2, wouldn't that give me a Vo of 0? I am just confused as how to incorporate that 1.00s time lag and such :(

    Also, because I've had much difficulty with these problems, can anyone give me some tips to solving these problems? Thanks so much!
     
  2. jcsd
  3. Sep 11, 2003 #2

    dduardo

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    Staff Emeritus

    you are confused by the time the second rock leaves. What you put says that the first rock reaches the water and then one second later you throw the second rock. Instead the first rock gets dropped, and while it is in the air, the second rock is thrown.

    Here is the detailed solution:

    For the first rock the equation is:

    y = y0 - v0*t -(1/2)*a*t^2

    y = 0m
    y0 = 43.9m
    v0 = 0m/s
    t= ?
    a = g

    0 = 43.9 - (1/2)*g*t^2

    Therefore the time it takes to get to the bottom is:

    t = 2.99s

    If the second rock left 1 second after, but both hit the water at the same time, the second rock only has 1.99s to reach the water

    Therefore the equation for the second rock is

    y = y0 - v0*t -(1/2)*a*t^2

    y = 0m
    y0 = 43.9m
    v0 = ?
    t= 1.99s
    a = g

    0 = 43.9 - v0*(1.99) - (1/2)*g*(1.99)^2

    you can now solve for v0 which is:

    v0 = 12.31 m/s
     
    Last edited: Sep 11, 2003
  4. Sep 12, 2003 #3

    HallsofIvy

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    Science Advisor

    Missrikku: Since DDuardo gave the full answer (very nicely), you might have missed seeing exactly where you made your mistake:

    The first stone takes 2.99 seconds to hit the water. You throw the second stone 1 second after the first and it hits the water at the same time as the first. That means the second stone falls for one second LESS than the first, not more! The "t" for the second stone is 2.99-1= 1.99 seconds NOT 2.99+1.
     
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