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Homework Help: Kinematics: How fast was he when he got off the ramp

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A bike is jumping between two ramps. The ramps are a height H , both with angles [tex]\theta[/tex] and seperated by a distance D. If he landed halfway down the landing ramp find the speed at which he left the launching ramp in terms of H, [tex]\theta[/tex] and D.

    2. Relevant equations

    1/2 * at^2 + vi*t + di = df
    (i'm sure you all know that one)

    3. The attempt at a solution

    Okay so here's what I did, but apparently it was completely off, but i don't know where to begin it. *see attachment*

    Attached Files:

  2. jcsd
  3. Dec 6, 2009 #2
    I like that eqn.

    We know the takeoff angle. D is straightforward, how do we express the landing point in terms of the variables given. We are told it is 1/2 way down the ramp, so what is df and di. What are the initial x and y velocities?

    How can we figure the time of flight?
  4. Dec 6, 2009 #3
    The landing point is:


    [tex]L[/tex] = [tex]\frac{H}{tan(\theta}[/tex]

    With the equation

    I separated all of it to x and y components and got:







    Once I have both of them, do i just have to square both of them, put them under the square route then, finish?


    then substituting for L
    Last edited: Dec 6, 2009
  5. Dec 6, 2009 #4
    I'll need a sec to look at this, just got back. Stay tuned.
  6. Dec 6, 2009 #5
    here is the approach I have taken, so far unfruitful cuz of the 1/2 ramp displacement which bungs up what is usually very clean algebra: let a = angle of the ramp.

    Delta x= V Cos(a)*t=.5H/tan(a) + d

    Delta y = 1/2H=1/2 gt^2+ V sin(a)*t (a quadratic in t but only the longer time will work)

    Normally what I do here would be substitute for t which leads to:

    let d+0.5H/tan(a)= C; then 1/2H=1/2 g (C/(V*cos(a)))^2 + (V sin(a)*C/V cos(a))
    " = " + C (tan(a))
    There may be a trig ID here which can be uses to simplify things considerably.
  7. Dec 6, 2009 #6

    so why did you substitute the t, and where did you get the C from?
  8. Dec 6, 2009 #7
    oh I see,

    so with the time it took in the x direction, you used that time in terms of v,d,h,theta to put it in the y equation, since that's the one where the time in air matters. So after all that, the only thing is solve for V?
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