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Urgent Dynamics Problem Help Needed.

  1. Aug 30, 2012 #1
    The Problem.
    12-101 Hibbler dynamics 13th edition.


    It is observed that the skier leaves the ramp A at an angle θA=15 degrees with the horizontal. if he strikes the ground at B, deterime his initial speed VA and the speed at which he strikes the ground (at B).

    http://s3.amazonaws.com/answer-board-image/d324abb1-8b50-49a3-a0dd-e2375f1be1a4.jpeg (Ignore the values not in the diagram itself)

    My issue:
    I understand how to get VA but, I have no idea how to the the speed at which he strikes the ground at B. Any help would be very appreciated.

    I do not even know where to start to find VB.
     
  2. jcsd
  3. Aug 30, 2012 #2

    Doc Al

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    Staff: Mentor

    Hint: How far below A is B?
     
  4. Aug 30, 2012 #3
    64m. I just dont know how to set up and re work the equations to solve for VB. I understand for VA its

    80=VA(cos15)t

    -64=VA(sin15)t-(1/2)(9.81)t^2

    where I would sub in VA=80/cos15(t) to solve for t and then plug t back into the first equation.
     
  5. Aug 30, 2012 #4

    Doc Al

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    Staff: Mentor

    Since you know VA, you can use either equation to solve for t. (Use the first one, of course.)

    How does Vy depend on time?

    (There are several ways to get the answer; one way just uses the distance, but since you have the time why not use it.)
     
  6. Aug 30, 2012 #5
    I know VA and t just not how to get VB. Doesnt Vy increase with time in this case as gravity acts downward?
     
  7. Aug 30, 2012 #6

    Doc Al

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    Right. What's the kinematic equation describing that relationship?
     
  8. Aug 30, 2012 #7
    Vy=(Vo)y+at ?
     
  9. Aug 30, 2012 #8

    Doc Al

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    Sure.
     
  10. Aug 30, 2012 #9
    But how does VB relate into that? Im guessing I need to find VBy and VBx and then do the

    sqrt of (VBy)^2 + (VBx)^2
     
  11. Aug 30, 2012 #10

    Doc Al

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    Staff: Mentor

    Exactly. VBx should be trivial.
     
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