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Projectile motion skateboarder help

  1. Oct 26, 2006 #1
    This problem is not fun:
    Code (Text):

    A skateboarder starts up a 1.0-m-high,
    30deg ramp at a speed of 7.0 m/s.
    The skateboard wheels roll without friction.
     
    I guees the first thing I'll ask is the 7 m/s at 30 deg the launch velocity and angle?
     
  2. jcsd
  3. Oct 26, 2006 #2

    Doc Al

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    Staff: Mentor

    No, 7 m/s is the speed at the bottom of the ramp. You'll need to figure out the speed when it takes off at the top.
     
  4. Oct 26, 2006 #3
    ok, I don't know how to resolve this. I understand that the skateboarder is going to be slower at the top and gravity is the only force stopping motion.
     
  5. Oct 26, 2006 #4

    Doc Al

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    If you want to find the speed at the top, use conservation of energy. Or find the acceleration up the ramp.
     
  6. Oct 26, 2006 #5
    I have not learned the conservation of energy yet. I just used the formula sqrt(2*g*sin(theta)*delta(x)
    which is
    sqrt(2*9.8*sin(30)*1.73=4.11 m/s^2

    I still get the answer wrong.
     
  7. Oct 26, 2006 #6

    radou

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    Use v = Vo + at, where Vo is your initial speed, and a the acceleration. Now, what does this acceleration equal while the skater is climbing up the ramp?
     
  8. Oct 26, 2006 #7
    I'm guessing -9.8? But I dont know the time so how can I use that?
     
  9. Oct 26, 2006 #8

    Doc Al

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    What's the 1.73 signify?

    What you are using here--and which is perfectly OK if done right--is the formula:
    [tex]v^2 = v_0^2 + 2 a \Delta x[/tex]

    To use it properly, answer these questions:
    What's the acceleration along the ramp?
    What's the displacement along the ramp?
    What's the initial velocity?
     
  10. Oct 26, 2006 #9
    Well honestly, I 've done this problem so much I really don't know anymore. I dont know if accelreation is 9.8 or -gsin(theta). The displacement I'm assuming is the hypotenuse which is 2.
     
  11. Oct 26, 2006 #10

    rsk

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    You're right, take g sin theta for the acc (remember it's negative) and the displacement along the ramp is 2.
     
  12. Oct 26, 2006 #11
    Ok so I got 6.26 m/s. Now the problem becomes projectile motion I hope. Now I just need to find the range.
     
  13. Oct 26, 2006 #12

    rsk

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    That's not what I get.....can you show me your calculation.

    As far as I can see, both your distance up the slope and the acceleration with have sin30 in them, so the sin30s will cancel.
     
  14. Oct 26, 2006 #13
    I did sqrt((7^2)+(2*-9.8sin(30)*2)).
     
  15. Oct 26, 2006 #14

    radou

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    Seems correct, as far as I'm concerned.. I did another 'method' too, to check. :smile:
     
  16. Oct 26, 2006 #15

    rsk

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    That gives me 49 - 19.6 which gives me 29.4. The sq rt of that is not 6.26
     
  17. Oct 26, 2006 #16

    radou

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    [tex]\sqrt{7^2 - 2\cdot9.81\cdot\sin(30)\cdot2} \approx 5.42[/tex]. :rolleyes:

    (...Which perfectly fits the expression Doc Al gave: [tex]v^2 = v_0^2 + 2 a \Delta x[/tex].)
     
  18. Oct 26, 2006 #17

    rsk

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    And which is why I said I didn't agree with 6.26 m/s
     
  19. Oct 26, 2006 #18

    radou

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    Ooops, sorry, I seem to have misread you. :shy:
     
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