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Conservation of Energy initial velocity

  1. Sep 28, 2008 #1

    I know if there was no initial velocity, the velocity at the bottom would just be sqrt(2gh), but I'm not sure how the initial velocity impacts the velocity at the bottom. I thought just adding the x-component of the velocity would be enough but apparently it's not :\
  2. jcsd
  3. Sep 28, 2008 #2
    Well if there was an initial velocity, then what was the initial kinetic energy? Just remember:

    [tex]KE_0 + PE_0 = KE + PE[/tex]
  4. Sep 28, 2008 #3


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    Welcome to PF!

    Hi vesperaka! Welcome to PF! :smile:

    Yes, do what chislam suggests. :smile:

    Another way of looking at it: your equation can be written (1/2)v12 = gh.

    So how would you put v0 into that?
  5. Sep 29, 2008 #4
    I understand that KE = 1/2mv^2, so when I tinker around with that I get

    V = sqrt (2KE/m)

    But that equation gives me a problem cause I can't just make KE a variable like that. I do know that all of the energy at the bottom is KE though. So:

    1/2mvi^2 + sqrt(2gh) = 1/2mvf^2

    When I take that equation and solve for Vf, I get this really long equation

    Vf = Vi^2 + [2 sqrt(2gh)] / m

    Is that the right answer? If I answer it incorrectly 1 more time I don't get credit so I'm being cautious, but to be honest I don't understand why it wouldn't work (unless I botched the algebra).
  6. Sep 30, 2008 #5


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    Hi vesperaka! :smile:

    No, you're very confused.

    KE = 1/2 mv^2. PE = mgh (not gh, and certainly not 2gh).

    And where did that sqrt come from? :confused:

    Try again! :smile:
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