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Conservation of Energy of a slide

  1. Oct 7, 2014 #1
    1. The problem statement, all variables and given/known data
    A child slides without friction from a height h along a curved water slide (Fig. P5.44). She is launched from a height h/5 into the pool. Determine her maximum airborne height y in terms of h and η. (Use q for η and h as appropriate.)
    p5_44.gif (Fig. P5. 44)

    2. Relevant equations
    I know you use conservation of energy to solve the problem but I just am completely lost on where to start at right now...
    which is: mgh = 1/2mv^2

    3. The attempt at a solution
    I don't know where to begin that's my problem..... could I just get some help getting started possibly???
     
  2. jcsd
  3. Oct 7, 2014 #2
    Here's a hint. This is a sequence to two separate problems.
    Problem 1. She slides from a height h to a height h/5. What is her launch speed?
    Problem 2. She is launched as a projectile with the speed determined in Problem 1 and at a launch angle theta. What is the maximum height of her trajectory?

    Chet
     
  4. Oct 7, 2014 #3
    Problem 1.) The launch speed would just be 0, correct? Or am I just totally missing something here.....
     
  5. Oct 7, 2014 #4
    You're missing something. If the girl's elevation decreases by a distance 4h/5, her potential energy decreases, so her kinetic energy (determined by her speed) must increase.

    Have you ever been down a water slide? If so, was your speed zero when you shot out the bottom of the slide?

    Chet
     
  6. Oct 7, 2014 #5
    Oh, the bottom..... whoops, I was thinking from the start....
    so h-h/5? Maybe?
     
  7. Oct 7, 2014 #6
    I don't understand what you're asking.
     
  8. Oct 8, 2014 #7
    Am I getting somewhere with (h-h/5)?
     
  9. Oct 8, 2014 #8

    Orodruin

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    That would depend on how you plan to use it ... It is the vertical distance between the top and end of the slide. What does this imply?
     
  10. Oct 8, 2014 #9
    Use the law of conservation of energy:
    [itex] mgh_1 + \frac{1}{2} mv_1 ^ 2 = mgh_2 + \frac{1}{2} mv_2 ^ 2 [/itex]
    At the top of the slide she has a velocity of zero, she starts at position 'h' and ends at position 'h/5', so solve for velocity and then apply the law of conservation of energy again, remembering that her horizontal velocity will be constant
     
  11. Oct 9, 2014 #10
    Phase 1, calculating the release velocity :
    Even though the girl hits the bottom of the slide then rises again, the crucial dimension is the height difference between the top of the slide and the exit point from the slide ( 0.8 h)
     
  12. Oct 9, 2014 #11

    NascentOxygen

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    You are getting warm. We're looking for the change in energy associated with that change in height.
     
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