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Kinda embarrassing to ask about setting up this simple kinematics eqtns

  1. Jun 21, 2011 #1
    1. The problem statement, all variables and given/known data

    I made this one up and now i can't solve it (tada...)

    [PLAIN]http://img577.imageshack.us/img577/5081/probloem1.jpg [Broken]

    OKay so the biker runs off the ramp and leaves the ramp at the tip with some initial velocity v0

    I will give some numerical values to make this easier

    ϕ = 30 degrees
    g = -10m/s2
    v0 = 20m/s
    y = 10m
    h = 20m
    d = 5m
    x = 30m

    OKay three goal

    1) Set up the motion equations for the biker
    2) Find the max height reached
    3) Find when it lands on top of the new building

    3. The attempt at a solution

    1) I have most problems with this one, especialyl when I cannot decide if ϕ is the correct angle. That is if the biker flies off from an angle different from ϕ

    [tex]y(t) = -5t^2 + |20|sin30t + (10 + 20)[/tex]

    [tex]x(t) = |20|cos30t[/tex]


    [tex]y(t) = -5t^2 + |20|sin30t + (10 + 20) = -5t^2 + 10t + 30 = -5(t^2 - 2t - 6) = -5(t^2 -2t + 1 - 1 - 6) = -5((t -1)^2 - 7) = -5(t-1)^2 + 35[/tex]

    So ymax = 35m

    3) So basically h - d = 20 - 5 = 15

    [tex]15 = -5^2 + 10t + 30[/tex]

    [tex] 0 = -5t^2 + 10t + 20[/tex]

    [tex]t = 1 + \sqrt{5}[/tex]
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 21, 2011 #2


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    Close, you should have a +15 in the second to last equation, not 20.

    Also, instead if figuring out the whole equation for a parabola with that silly math you did, you can immediately figure out when the biker reaches the top of the parabola because you know [itex]v_{y}(t) = v_0 - at[/itex] and you simply set the velocity to 0 and you know immediately what time you reached the top.
  4. Jun 21, 2011 #3
    It's called being lazy...

    Just wondering why is using phi right? I thought it should have left with a different angle than phi
  5. Jun 21, 2011 #4
    At the instant that the biker leaves the ramp what is the angle of the velocity vector? that is whi phi is the right angle to use. Also, why do you think that it should be some other angle?
  6. Jun 21, 2011 #5
    What if he doesn't actually fly off? But instead trips with that velocity?
  7. Jun 21, 2011 #6


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    Well that's like saying what if he gets hit by a meteor instead of flying off.

    He's going to fly off at that angle because there is no other angle he could possibly fly off at due to inertia
  8. Jul 5, 2011 #7
    Actually something still doesn't make sense, it came to me last night.

    [tex]y(t) = -5t^2 + |20|sin30t + (10 + 20)[/tex]

    This says that the biker is subjected to gravity even before he leaves the ramp, but when he is on the ramp, there is a normal force, so he isn't free-falling like the equation says. This equation is only valid when he leaves the ramp no?
  9. Jul 5, 2011 #8


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    The equation is only valid after the person has left the ramp.

    If you want to calculate the equations of motion while he's on the ramp, he wouldn't be in free fall and you'd have to take into account the force he exerts to push himself up the ramp and the normal force and all that.
  10. Jul 5, 2011 #9
    But that equation is a parabolic path, how do I know at what value of t does the equation become valid?
  11. Jul 5, 2011 #10


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    By convenience you start the problem at t=0. Whatever happens when the guy is on the ramp is taken care of by the fact that you know the end velocity and the angle he flies off at.
  12. Jul 5, 2011 #11
    I just graphed it and it was so consistent that I am in disbelief.
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