1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

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

    With the equation
    [tex]\frac{1}{2}at^{2}+v_{i}t+d_{i}[/tex]

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

    [tex]v_{i}t=-\frac{at^{2}}{2}+d_{f}[/tex]
    then

    [tex]v_{i}=-\frac{at^{2}}{2t}+\frac{d_{f}}{t}[/tex]

    so

    [tex]v_{i}_{x}=-\frac{at}{2}+\frac{D+\frac{L}{2}}{t}[/tex]

    and

    [tex]v_{i}_{y}=-\frac{gt}{2}-\frac{H}{2t}[/tex]

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

    [tex]v_{i}=\sqrt{(-\frac{at}{2}+\frac{D+\frac{L}{2}}{t})^{2}+(-\frac{gt}{2}-\frac{H}{2t})^{2}}[/tex]

    then substituting for L
    [tex]v_{i}=\sqrt{(-\frac{at}{2}+\frac{D+\frac{(\frac{H}{tan(\theta)})}{2}}{t})^{2}+(-\frac{gt}{2}-\frac{H}{2t})^{2}}[/tex]
     
    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
    Thanks,

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook