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Projectile Motion with different starting and ending heights

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data

    A motorcyclist is going off an incline 53 degrees above the horizontal. He wants to land on a ledge 40 meters from the ledge he is launching off of, 15 meters below the edge of the ramp he is laughing off of. Δx = 40m. Δy = -15m. What is the minimum initial velocity he needs to go off the ramp at to barely make the jump?

    2. Relevant equations

    Vx = Vocos53
    Ag = 9.8 m/s^2
    Voy = Vosin53
    x = Vx(t)
    t = ?
    y = yo + Voy(t)-1/2(a)(t)^2


    3. The attempt at a solution

    If this was a horizontal launch it'd be very simple, but I have no idea what to do with launching off an incline. I tried:

    t = square root (2d/a) = square root ((2*15)/9.8) = 1.75 sec

    x = Vx(t) therefore 40 = Vx(1.75) ; Vx = 22.86 m/s

    Then; Vo = Vx/cos53 = Vo = 22.66/cos53 = 37.99 m/s. The book says this is wrong and I have no idea what to do.
     
  2. jcsd
  3. Feb 21, 2010 #2

    ideasrule

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    Homework Helper

    t does not equal the square root of 2d/a because the initial vertical speed is not 0. Do you know which equation 2d/a came from? Use that equation; you now have an equation relating vertical displacement (15 m), time, and initial speed. There's one other equation that you need, and it's relating to horizontal speed.
     
  4. Feb 21, 2010 #3
    Unfortunately, that leaves 2 unknowns in 1 equation. The unknowns being initial speed and time.
     
  5. Feb 21, 2010 #4

    ideasrule

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    I said that you need one other equation--namely, the one relating to horizontal speed.
     
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