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Basic Motion Problem (Dropping Objects)

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data

    You are given a length of rope, tape and five rocks. The rocks must be attached in such a manner that when the rope is released, the sound of each rock hitting the ground is evenly spaced.

    There must be a rock at the very bottom and the very top of the rope. When the rope is released, the bottom rock will be just above the ground. A stopwatch cannot be used (time cannot be recorded).

    2. Relevant equations

    Any of the following kinematics equations:

    d=1/2(Vf+Vi)t
    Vf=Vi+at
    d=Vit+1/2at^2
    d=Vft-1/2at^2
    Vf^2=Vi^2+2ad

    3. The attempt at a solution

    a.) I understand that the rope will accelerate downwards due to gravity. This means that the rock at the top of the rope will have a faster final velocity just before impact than the rock at the bottom of the rope.
    b.) Assuming that the bottom rock takes 0.1s to hit the ground I made the following calculations:

    (Trying to space the sounds 0.2 second apart):

    2nd rock from the bottom:

    d = Vit+1/2at^2
    d = (0)t + 1/2(9.81m/s^2)(0.3s)^2
    d = 0.44 m

    This means the second rock will be placed 0.44 m from the first rock (0.44 m from the bottom of the rope).

    3rd rock from the bottom:

    d = Vit+1/2at^2
    d = (0)t + 1/2(9.81m/s^2)(0.5s)^2
    d = 1.23 m

    This means that the third rock will be placed 1.23 m from the bottom of the rope.


    Is this the correct approach?
     
  2. jcsd
  3. Sep 16, 2012 #2

    CWatters

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    Science Advisor
    Homework Helper

    Close but you are given the length of rope so the distances have to add up to that length. Hard to do it that way.

    Perhaps write equations for dtop and dbot in terms of Δt. where Δt = one time interval.

    The length of the rope is then dtop - dbot. Rearrange to give Δt ?

    Something like that.
     
    Last edited: Sep 16, 2012
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