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Physics- Mechanical Energy

  1. Apr 24, 2003 #1
    An arrow is shot from ground (height=0) at a speed of 46m/s and travels in an archy line and at the maximum height has a speed of 42m/s. Arrows mass is, m...what is the maximum height that the arrow reaches? **IGNORE ANY FRICTION**

    Thanks for any help guys


    What I did was:

    Et1=Mechanical energy before arrow was shot
    Et2=Mechanical Energy after arrow was shot
    Vf2=Final velocity squared
    Vf1=Initial velocity squared



    Et1=Et2

    m(vf2)/2 - m(vi2)/2 = mgh + m(vf2)/2 - m(vi2)/2

    ...it doesnt work out though...why???
     
    Last edited: Apr 24, 2003
  2. jcsd
  3. Apr 24, 2003 #2

    Tom Mattson

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    You've got the right idea.

    You've double-counted the kinetic energies. Each one should only appear once.

    The total initial energy is Ei=(1/2)mvi2

    The total final energy is Ef=(1/2)mvf2+mgh

    Just equate and solve.
     
  4. Apr 24, 2003 #3
    But why though, I dont get it?
     
  5. Apr 24, 2003 #4

    Tom Mattson

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    What don't you get? The expressions for the total energies, or the mistake of double counting kinetic energy?
     
  6. Apr 24, 2003 #5
    Double counting the Kinetic energy thing....
     
  7. Apr 25, 2003 #6

    Tom Mattson

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    Re: Re: Physics- Mechanical Energy

    Continuing with what I wrote earlier:

    Equate the energies Ei=Ef

    (1/2)mvi2=(1/2)mvf2+mgh

    See? You put in an extra -(1/2)mvi2 on the RHS and an extra -(1/2)mvf2 on the LHS.
     
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