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Homework Help: Sled Jump

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Cal and Henry are riding on a sled. They are trying to jump the gap between two symmetrical ramps of snow separated by a distance W . Each ramp makes an angle θ with the horizontal. They launch off the first ramp with a speed VL. Cal, Henry and the sled have a total mass m.

    Find the initial launch speed VL that will result in the sled landing exactly at
    the peak of the second ramp. Express your answer in terms of some (or all) of the parameters m, θ, W, and the acceleration of gravity g, Include in your answer a brief description of the strategy that you used and any diagrams or graphs that you have chosen for solving this problem. Make sure you clearly state which concepts you plan to use to calculate any relevant physical quantities.

    2. Relevant equations

    3. The attempt at a solution

    y = yo + vyt - .5gt2

    I set y = yo because they first & second ramp are the same, so they cancel out

    0 = vyt - .5gt2

    v = [tex]\frac{4.9t}{sin\theta}[/tex]

    the answer is this .... v = (gW / sin 2θ)1/2

    I cant understand how they get that, can i get a hint
  2. jcsd
  3. Oct 1, 2009 #2


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    Homework Helper

    As in all trajectory problems you must find the vertical and horizontal components of the initial velocity. Then make two headings
    Ask yourself whether there is constant speed or accelerated motion in each case and write the appropriate formula(s). Fill in the information you have and solve one of the formulas for time. Sub the time into the others to get the info you want.
  4. Oct 4, 2009 #3
    for the x direction

    x = xo + vxt

    x = 0 + vcos[tex]\theta[/tex]t

    t = x / (vcos[tex]\theta[/tex])

    v = gt/2sin[tex]\theta[/tex] = g(x / (vcos[tex]\theta[/tex]))/2sin[tex]\theta[/tex] = gx / 2vsin[tex]\theta[/tex]cos[tex]\theta[/tex] = (gx/sin2[tex]\theta[/tex]).5

    i think i got it
  5. Oct 4, 2009 #4


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    Homework Helper

    The first 3 lines look good, though x should be replaced by w. I don't know what that v= line is about.
    For the y direction, you should write two equations:

    d = Vi*t + .5*a*t^2 and V = Vi + a*t
    and put your numbers into both.
  6. Oct 5, 2009 #5
    the v is from my original post, and then i proceded to substitute in the t, I got the correct answer, thanks for ur help
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