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Kinematics of jet-ski - how far does Sam land?

  1. Jun 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Sam (5) kg takes off up a 50-m high, 10 degree frictionless slope on his jet-powered skis. The skis have a thrust of 200 N. He keeps his skis titled at 10 degrees after becoming airborne, as shown in the Figure. How far does Sam land from the base of the cliff?

    I drew the picture on paint. here is the link for it:
    http://i196.photobucket.com/albums/aa59/aliatehreem/Sam.jpg


    2. Relevant equations

    F= ma
    vf^ 2 = vi ^2 + 2ad
    vx= v* cos theta
    vy= v* sin theta

    yf= yo + vt+ 0.5 *a* t^2

    3. The attempt at a solution

    Find length of slope:

    h= 50/ (sin 10)

    First I have the acceleration parallel to the motion.
    F= ma
    ma= F (thrust) - weight parallel
    ma= 200 - mg (sine theta)
    a= 0.96318

    Find velocity at the end of the slope.

    vf^ 2 = vi ^2 + 2ad
    vf^ 2 = 0 + 2(.96318)(50/ (sin 10) )
    vf = 23.55 m/s

    This velocity is parallel to the slope.

    To find velocity after Sam gets off the slope, in the vertical and horizontal direction, the x and y component velocities are:

    vx= v* cos theta= 23.55* cos10= 23.2
    vy= v* sin theta= 23.55* sin 10= 4.09

    Since vertical displacement is -50 m, find time, and sub into horizontal component.

    yf= yo + vt+ 0.5 *a* t^2
    -50= 0 + 4.09 t + 0.5 * -9.81 * t^2
    t= 3.637 s

    Sub into horizontal velocity to find distance:

    xf= xo + vt
    xf = 23.2 (3.637) = 84.4 m

    The correct answer is 110 m. Can someone please tell me what I am doing wrong? Thanks in advance!
     
  2. jcsd
  3. Jun 18, 2008 #2
    First I have the acceleration parallel to the motion.
    F= ma
    ma= F (thrust) - weight parallel
    ma= 200 - mg (sine theta)
    a= 0.96318

    Check these calculations. I think this is where your mistake is.
     
  4. Jun 18, 2008 #3
    That was a typo. Sorry. Sam's mass is actually 75 kg.
     
  5. Jun 18, 2008 #4
    ah. I knew something was off. Hold on
     
  6. Jun 18, 2008 #5

    rl.bhat

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    ma= 200 - mg (sine theta)
    a= 0.96318

    Check this calculation
     
  7. Jun 18, 2008 #6
    I'm getting the same answer:

    ma= 200 - mg (sine theta)
    75a= 200 - (75)(9.81) (sin 10 degrees)
    a= 0.96318 m/s^2.
     
  8. Jun 19, 2008 #7

    rl.bhat

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    Your answer appears to be correct.
     
  9. Jun 19, 2008 #8
    For the acceleration, you mean? The overall answer for some reason isn't turning out to be correct. I checked my calculations many times. In the beginning, I made a few error but then I corrected those, but the answer I'm supposed to be getting is 110 m, not 84.4 m :(. I hate textbook questions because they don't even have proper solutions, they just give the final answer without any explanation!
     
  10. Jun 19, 2008 #9

    Kurdt

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    Sam is still subject to some thrust after he leaves the ramp if I'm reading the question correctly.
     
  11. Jun 19, 2008 #10
    Thanks a lot! That solves the mystery. I assumed that the force of thrust was only present while on the ramp. So now, I have an acceleration in both x and y directions and I can use a similar method to solve for the horizontal distance.
     
  12. Jun 19, 2008 #11

    Kurdt

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    Well tell me that when you get the right answer. I just had a quick look. :tongue:
     
  13. Jun 19, 2008 #12

    I got the answer now! Yay! Thank you so much for guiding me! I love this forum! I'm feeling more confident now and learning more things by second :D.
     
  14. Jun 19, 2008 #13

    Kurdt

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    Very good! I hope you'll be around helping other students one day. :smile:
     
  15. Jun 19, 2008 #14
    You never know!
     
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