Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Ski Jump (Conservation of Energy)

  1. Oct 10, 2012 #1
    1. The problem statement, all variables and given/known data
    A skier (m=55.00 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. (A)If h = 6.80 m and D = 10.20 m, find H.(B)Find her total kinetic energy as she reaches the ground.


    2. Relevant equations
    KEi+PEi=KEf+PEf (law of conservation of energy)
    KE=1/2mv^2
    PE=mgh

    3. The attempt at a solution
    I somehow have to find the total Height and the final Velocity. I can't seem to come up with an equation that doesn't have one of those in it.....Any help would be greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Oct 10, 2012 #2
    You can find the velocity at the bottom of the ramp with the information given. You can treat it like a projectile problem, where the skier leaves the ramp at 90 degrees, with an initial height h, and lands a distance D from the ramp. Once you find this velocity, you can use the energy equations that you posted above to find H.
     
  4. Oct 10, 2012 #3
    omg...I don't see how I didn't see it before....Thank you very much!
     
  5. Oct 10, 2012 #4
    Wait....Don't you need to know how fast it's going off the ramp?
     
  6. Oct 10, 2012 #5
    First, you find the time it takes for the skier to fall height h (this is independent of the speed of the skier, since the ramp is horizontal). Then, you can find how fast the skier must have been going to travel a distance D in this time.
     
  7. Oct 10, 2012 #6
    So I got V=D/sqrt(2*h/g)
    Then plug that into: m*g*H=1/2*m*(D/sqrt(2*h/g))^2 Since H=x+h we get
    m*g*(x+h)=1/2*m*(D/sqrt(2*h/g))^2 then solve for x giving us:
    x=((.25*d^2)/h)-h Does that seem correct?

    I got -2.975???

    Never mind I forgot to add in the mgh to the end of the equation silly me thanks again for all your help:)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook