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Homework Help: Claculate the final speed of a skier who skis down a hill

  1. Mar 8, 2005 #1
    I have an assignment question, it states: Surprisingly little advantage is gained by getting a running start in a downhill race. To demonstrate this, claculate the final speed of a skier who skis down a hill 80-m-high with negligible friction (a) if her initial speed is zero; (b) if her initial speed is 3.0 m/sec. [this final speed found in part (b) is larger than in part (a), but by far less than 3.0 m/sec!]
  2. jcsd
  3. Mar 8, 2005 #2
    Use this kinematic equation:

    [tex]v_{f}^2 = v_{i}^2 + 2ad[/tex]

    Compare the two.
  4. Mar 8, 2005 #3


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    For both cases:
    {Delta Kinetic Energy} = -{Delta Potential Energy} = mgh = (9.81)(80)*m = (785)*m =
    = m*(vf)2/2 - m*(vi)2/2 =
    = (m/2)*{(vf)2 - (vi)2}
    ::: ⇒ (vf)2 - (vi)2 = (1570)
    ::: ⇒ vf = sqrt{1570 + (vi)2}

    Case #1:
    vi = 0 ::: ⇒ vf = sqrt{1570 + (0)2} = (39.6 m/sec)

    Case #2:
    vi = (3.0 m/sec) ::: ⇒ vf = sqrt{1570 + (3)2} = (39.7 m/sec)

    Last edited: Mar 8, 2005
  5. Mar 8, 2005 #4
    Thank you
    xanthym, your response is a little confusing.. is that potential energy that you're using?

    Jameson, i don't get how i'd get a? also would distance be 80 m? and if so isn't that height or are they the same thing?
  6. Mar 8, 2005 #5
    xanthym's right. I just glanced at this question... you don't have enough information to use a kinematic equation.

    You know the potential energy at the top, and since their is no kinetic energy yet, you can call this total the total energy.

    [tex]P.E. = mgh[/tex]
    [tex] = (m)(9.81)(80)[/tex]

    It is interesting to note that her mass does not affect this problem, as it will cancel out because it is used in every term.

    Look at xanthym's work, and see how using the Law of Conservation of Energy will help you figure out the final velocities.

  7. Mar 8, 2005 #6
    Ok so then
    its PE=gh since mass cancels out?

    this is what i have

    KEi=PEi =KEf + PEf
    becomes 0 + mgh inital = 1/2mv final squared + 0

    v final=mgh inital = 2gh initial

    v final= (2gh inital) 1/2

    sorry i hope that makes sense.. i'm not sure how to use the latex thing..
  8. Mar 8, 2005 #7
    [tex] (2gh + v_i^2)^\frac{1}{2}[/tex]
    Last edited: Mar 8, 2005
  9. Mar 8, 2005 #8
    okay i got it =)

    thank you so much!
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