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2-D Kinematics

  1. Sep 15, 2006 #1
    The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply upward at the end. The sharp upward slope launches them into the air, where they perform acrobatic maneuvers. In the women's competition, the end of a typical launch ramp is directed 63° above the horizontal. With this launch angle, a skier attains a height of 11.7 m above the end of the ramp. What is the skier's launch speed?

    i dont know where to begin with this since so little info is given,
     
  2. jcsd
  3. Sep 15, 2006 #2

    NateTG

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    This is mostly a 1-D problem.
     
  4. Sep 15, 2006 #3
    ok, i have the 3 sides of the traingle, but i do now know what to do with them, 13.13, 11.7, 5.96
     
  5. Sep 15, 2006 #4

    NateTG

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    Here's a slightly easier problem you could solve first:
    If I throw a ball straight up, how fast do I need to throw it so that it goes up 11.7 meters?
     
  6. Sep 15, 2006 #5
    EDIT: u are right it is too high.. 15.14m/s is the right one i think
     
    Last edited: Sep 15, 2006
  7. Sep 15, 2006 #6

    NateTG

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    That number seems a bit high ....

    Now, let's say that I throw the ball at a 45 degree angle with horizontal instead of straight up - how fast do I have to throw it?
     
  8. Sep 15, 2006 #7
    i have no idea
     
  9. Sep 15, 2006 #8
    can anyone PLEASE help, i have the answer(16.99), i just dont know how to get the answer
     
  10. Sep 15, 2006 #9

    NateTG

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    What's the vertical component of the velocity of the ball when I throw it?
     
  11. Sep 15, 2006 #10
    10.7m/s ??
     
  12. Sep 15, 2006 #11

    NateTG

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    It would only go up about 5m if that was the vertical component.
     
  13. Sep 15, 2006 #12
    Vy of the ball at 45degrees or Vy of the ball thrown straight up??? wow if every problem is going to take me this long i think ill be done next month
    can you give me a forumla or something for the actual problem
     
    Last edited: Sep 15, 2006
  14. Sep 15, 2006 #13
    anyone? im completely lost
     
  15. Sep 15, 2006 #14

    NateTG

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    The way I would do this problem is:
    1. Figure out the vertical component of the skiers velocity at the end of the ramp.
    2. Use trig to determine the total velocity.
     
  16. Sep 15, 2006 #15
    How would i find the vertical component of the skiers velocity, when it does not give the velocity

    IS 11.7 the vertical component?
     
    Last edited: Sep 15, 2006
  17. Sep 15, 2006 #16

    NateTG

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    The horizontal and vertical components are independant.
     
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