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One Dimensional Question

  1. Dec 29, 2011 #1
    1. When a kid drops a rock off the edge of a cliff, it take s 4.0s to reach the ground below. When he throws the rock down, it strikes the ground in 3s. What inital speed did he give the rock?


    ΔX=ViTf + 1/2ATf^2
    H=1/2gt^2
    Vf^2=Vi^2+ 2AΔX

    3. Hi everyone! :smile: I know that when the kid drops the rock, you can use the free fall equation- H=1/2gt^2; but the problem doesn't say the height of the cliff and I don't know "g", unless it's just 9.8? :confused: Also, I don't know inital velocity, so I just don't know how to go about this! If anyone can help me, that'd be much appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 29, 2011 #2

    SammyS

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    Assuming the cliff is on the earth, then yes, g = 9.8 m/s2.


    When it says the kid drops the rock, you can assume that the initial velocity is zero.
     
  4. Dec 29, 2011 #3
    Hmm...why would there be two different scenarios though? One with the kid dropping the rock, and the other him throwing it? And I don't understand what the question is asking with "intial speed", is that just inital velocity? Then wouldn't they both be zero?
     
  5. Dec 29, 2011 #4

    gneill

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    Because that is part of the problem that is being posed.
    When you throw something it's initial speed isn't zero. It's whatever speed it leaves your hand.

    Also, velocity is just a fancy speed: it has magnitude and direction :smile: In this case you can take the initial speed to be directed downwards and call it the initial velocity if you like.

    I suggest that you use the first scenario (rock is dropped) to find the height of the cliff. Then use the first of your equations to wrangle with the second scenario.
     
  6. Dec 29, 2011 #5
    Good advice!

    Does this look right?

    H=1/2gt^2
    H=1/2(9.8)4s
    H=19.6m

    ΔX=ViTf + 1/2 ATf^2
    19.6=Vi (4) + 1/2(o)Tf^2
    Vi=4.9m/s

    Thanks:smile:
     
  7. Dec 30, 2011 #6

    gneill

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    Whoops. You forgot to square the time; that's a t2 up there.
     
  8. Dec 30, 2011 #7
    Ahh, I gotcha. Were those the right steps to solve it though?
     
  9. Dec 30, 2011 #8
    Wait...does the 3s from when he throws the rock ever come into play?
     
  10. Dec 30, 2011 #9

    gneill

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    Of course. It takes 3s to hit bottom when he throws it rather than drops it. That's the effect of the initial velocity.
     
  11. Dec 30, 2011 #10
    So in the ΔX=ViTf+1/2ATf2 the Tf would be 3s not 4s?
     
  12. Dec 30, 2011 #11

    gneill

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    Do you think it would take the same time for both scenarios?
     
  13. Dec 30, 2011 #12
    No..so do this problem twice to find his inital speed for both scenarios, using 4s for the first and 3s for the second?
     
  14. Dec 30, 2011 #13

    gneill

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    When you DROP an object its initial speed is zero. The initial scenario allows you to calculate the height of the cliff. Use this height in the second scenario.
     
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