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Homework Help: Snowboarding trick problem

  1. Jun 16, 2011 #1
    1. The problem statement, all variables and given/known data
    How much height above the half pipe does 85kg Shaun White need to complete the 3.5 second trick?

    2. Relevant equations
    net force= ma
    vf= vi+at

    3. The attempt at a solution
  2. jcsd
  3. Jun 16, 2011 #2
    What kinematics equations do you know? What information does the problem give you, and is there anything else you know? You must try to answer the question before we can help.
  4. Jun 16, 2011 #3
    You can solve for his exit-velocity using one of your listed equations and the relationship between [itex]v_f[/itex] and [itex]v_i[/itex], since acceleration and time are both known.

    Then use one of the kinematic equations (perhaps one that relates distance to velocity and acceleration? :wink:) to find how high he goes.
  5. Jun 16, 2011 #4
    What is the acceleration?
  6. Jun 16, 2011 #5
    What pulls him back down into the halfpipe? :)
  7. Jun 16, 2011 #6
    oooohhhh ookk... Thanks!
    so it would be
  8. Jun 16, 2011 #7
    Almost! But not quite. His initial velocity will be the velocity he has when he leaves the halfpipe, so it will not be zero. What's the relationship between the velocity at which he leaves the halfpipe and the velocity at which he re-enters the halfpipe? (Hint: Assume air resistance is negligible and energy is conserved)

    Also consider using a different kinematic equation to find the distance (we have velocity and acceleration and want to find his MAXIMUM distance).
  9. Jun 16, 2011 #8
    is the initial velocity 34.3m/s?
  10. Jun 16, 2011 #9
    If nothing acts on him except for gravity when he's doing his trick, what can we conclude about his velocity upon re-entering the halfpipe in terms of the velocity when he leaves the halfpipe?
  11. Jun 16, 2011 #10
    that his velocity is 9.8m/s
  12. Jun 16, 2011 #11
    We're not looking for a number for [itex]v_i[/itex] yet -- we'll solve for that later. We want to find a relationship between that and [itex]v_f[/itex]. Does this equation hold true:

    [itex]v_f = v_i[/itex]?

    Well of course not, because velocity has a direction and magnitude. The directions clearly are opposite, and what can we say about the magnitudes?
  13. Jun 16, 2011 #12
    so vf =0?
  14. Jun 16, 2011 #13
    I think I'm doing a bad job at explaining this, so I'll help you out. If all energy is conserved, then at the height of the exit of the halfpipe between [itex]t = 0[/itex] and [itex]t = 3.5[/itex],

    [itex] \Delta E = 0[/itex]

    And thus,

    [itex]\Delta KE = \frac{1}{2} m(v_f^2 - v_i^2) = 0 [/itex]

    Which, as a result, gives us:

    [itex]|v_f| = |v_i|[/itex]

    The only difference is that [itex]v_i[/itex] is going the opposite direction, so:

    [itex]v_f = -v_i[/itex]

    Now try to solve for [itex]v_f[/itex] numerically using your kinematic equation.
    Last edited: Jun 16, 2011
  15. Jun 16, 2011 #14
    ok, i think i got it, thanks for your help!
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