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How far does Sam land from the base of the cliff?

  1. Oct 14, 2009 #1
    Sam (60 kg) takes off (from rest) up a 50 m high, 10° frictionless slope on his jet-powered skis. The skis have a thrust of 200 N. He keeps his skis tilted at 10° after becoming airborne, as shown in Figure P6.43. How far does Sam land from the base of the cliff?

    a=F/m


    I used 50/sin10 to get the length of the slope-287.94m. Then i calculated the f on the slope. 200N-mgsin10. then i use that to divided by the mass which gives 1.63 for a. after that i go the final velocity before it jump off. then i have no idea wat to do next
     
  2. jcsd
  3. Oct 14, 2009 #2

    ideasrule

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    How long does he stay airborne for? How far, horizontally, does he travel in that time? If you don't know, look at your equations for kinematics and projectile motion.
     
  4. Oct 14, 2009 #3
    i know i need to get the time from the y axis then calculate the distance on x axis. but i am not sure wat do i put for Vi in the y axis
     
  5. Oct 14, 2009 #4

    ideasrule

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    I thought you already calculated the speed that Sam leaves the cliff at. Just do Vsin10 to find the vertical component of his speed and that's your Vi.
     
  6. Oct 14, 2009 #5
    so t=-50=(30.6528sin10)t+.5(-9.8)t^2
    t=3.78
    then for the x axis
    (30.6528cos10)3.78+.5(200/60)(3.78^2)
    the acceleration on x axis is just 200/60 right
     
    Last edited: Oct 14, 2009
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