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Question help

  1. Nov 30, 2005 #1
    A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As Figure 3 below shows, a 50 kg boy is observed to hit the water 7 meters from the end of the slide, as shown. The slide is frictionless. Also, ignore air resistance.
    Find the speed, vB ,at which the boy leaves the slide, the initial height, H, of the boy, and the speed, vC, of the boy upon hitting the water.

    I used the eqaution 1/2 mvi^2+mgyi=1/2 mvf^2 +mgyf to find the final velocity of the boy but I dont have the yi (initial height)

    If you can tell me what to figure out first to get me going and how to do it I know I can figure out the rest.
     
  2. jcsd
  3. Nov 30, 2005 #2

    mezarashi

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    The steps are:

    1. Find the velocity of the boy as he leaves the horizontal end of the slide.
    2. Find the time it takes for him to hit the water
    3. Using that time, find the vertical distance he must have fallen from (i.e. how high is the slide above the water).
     
  4. Nov 30, 2005 #3
    yi (the initial height of the boy) is unknown and I cant figure out the velocity of the boy as he leaves the slide at the horizontal without it, how would I go about figuring it out
     
  5. Nov 30, 2005 #4

    Fermat

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    Have you given all the information ?

    Do you know the height of the end of the slide above the water level ?

    I've not been able to use the weight of the boy, which makes me wonder if all the info has been given.
     
  6. Nov 30, 2005 #5
    yeah actually the height at the end of the slide is 2.4m ( i forgot to mention it)
     
  7. Nov 30, 2005 #6

    Fermat

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    Then it all falls out :smile:

    Have you got any work done, any equations ?

    Oh yes, I see you've got something, but can you say where vf and yf apply to ? Is that the velocity at the end of the slide, or upon hitting the water ?
     
  8. Nov 30, 2005 #7
    the vf that I was trying to calculate was the velocity of the boy as he comes off the slide and yf would be 2.4
     
  9. Nov 30, 2005 #8

    Fermat

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    OK, that's one equation, in vf, yf and yi where yi and vf are the two unknowns.

    Now analyse the motion of the boy after he has left the end of the slide (with horizontal velocity vf) as mezarashi detailed above.
     
  10. Nov 30, 2005 #9
    so...
    the equation for the two unknowns (yi and yf) will be
    1/2(50(0)^2+ 50(9.8)yi=1/2(50)vf^2+(50)(9.8)(2.4)
    490yi=25vf^2+1176

    and to analyze the motion of the boy at the end of the slide
    would you use the same equation but use yi as 2.4 and yf as zero becuase he is now in the water?
     
  11. Nov 30, 2005 #10

    Fermat

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    You used yf before as the height of the end of the slide. It would be inconsistent (and bad form :frown: ) to use yf to denote another dimension.
    Why not use yw (y at the water) as the height above the water, with yw = 0, of course. And keep yf = 2.4 m.

    Also, you don't use the same eqn. That was the eqn for conservation of energy. You should use the kinematic eqns of motion for the travel of the boy during this period.
     
  12. Nov 30, 2005 #11
    so when using the kinematic equation i would be delta y as 2.4 and solve for v but wouldn't I need to know the acceleration how owuld I do that
     
  13. Nov 30, 2005 #12

    Fermat

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    The only acceleration involved is that of gravity.
     
  14. Nov 30, 2005 #13
    so a would be 9.8?
     
  15. Nov 30, 2005 #14

    Fermat

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    yep, that's it.
     
  16. Nov 30, 2005 #15
    could I use a projectile motion equation vy^2 = voy^2 - 2g(delta y)?
     
  17. Nov 30, 2005 #16
    no wait Ineed to find the time so thats not the right equation
     
  18. Nov 30, 2005 #17

    Fermat

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    You know how high he is (2.4m). How long would it take for him to fall that distance under gravity ?
    If his horizontal velocity is vf, how long would it take him to travel 7m (horizontally) ?
    Equate the two times
     
  19. Nov 30, 2005 #18
    deltay = voy t + 1/2 ay t^2 is it this one? where t=0.6998
     
  20. Nov 30, 2005 #19

    Fermat

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  21. Nov 30, 2005 #20
    deltay = voy t + 1/2 ay t^2 y direction
    deltax = vox t + 1/2 ax t^2 x direction

    did i chose the right eqautions
     
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