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Conservation of Energy Problem

  1. Jun 15, 2005 #1
    Here's a problem involving conservation of energy. My answer is wrong, but I can't figure out why. I would be extremely grateful for some help :smile:

    A skier (m=59.00 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h, the height of the end of the ski jump, is 3.40m and D, the distance from the end of the jump to his landing point, is 10.90 m, find H, the height of the top of the ski jump.

    The answer I'm getting is 55.63m

    Step 1: To begin, I calculated the horizontal velocity of the skier at the end of the jump.
    v^2=2gm
    v^2=2(9.8m/s^2)(59kg)
    v=34.01m/s
    Step 2: Then, I calculated the time, using Newton's Kinematic Equations for Projectile Motion.
    y=y0+vy0t-1/2gt^2
    -3.4=0+0-1/2(9.8)t^2
    t=.833s
    Step 3: I calculated the x and y components of the velocity of the skier when he lands on the ground.
    vx=vx0
    vx=34.01m/s
    vy=vy0-gt
    vy=0-9.8(.833)
    vy=-8.16m/s
    Step 4: I calculated the velocity of the skier when he lands by using the Pythagorean Theorum.
    v=33.02m/s
    Step 5: I used the equation for conservation of energy to determine the height at the top of the ski jump.
    KE1+PE1=KE2+PE2
    KE1=0
    PE1=mgy1=(59kg)(9.8m/s^2)(y1)=578y1
    KE2=1/2mv^2=1/2(59kg)(33.02m/s)^2= 32164.45J
    PE2=mgy2=(59kg)(9.8m/s)(0)=0
    0+578.2y1=32164.45J+0
    y1=55.63m
     
  2. jcsd
  3. Jun 15, 2005 #2
    I'll bet your problem is that your first equation is wrong. The units don't match. you have m^2/s^2=kg*m/s^2. The equation you want is v-final^2=v-initial^2+2gh, and I also believe you mean vertical velocity, not horizontal in your first step.
     
  4. Jun 15, 2005 #3
    That first equation isnt right, if you're using energy analysis, that m should be an h.

    This is a really unclear problem. If he takes off horizontally, his max height is 3.40m
     
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