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Projectile Motion Given only angle and distance?

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data

    A person decides to jump a canyon. The walls are equally high and 10 meters apart. He takes off by driving a motorcycle up a short ramp sloped at 11 degrees. What minimum speed must he have to clear the canyon?

    2. Relevant equations

    I don't know if these are correct or not

    (Time in Air)
    h= Vyt + 1/2gt[itex]^{2}[/itex]

    (distance)
    d= Vxt

    Vx= Vcos11
    Vy= Vsin11

    Other Formulas
    V[itex]^{2}[/itex]f= V[itex]^{2}[/itex]i + 2ad


    3. The attempt at a solution

    my sad attempt lol

    Vi=?
    Vf=?
    h=?
    d= 10 meters
    angel= 11 degrees
    g(or a)= 9.8m/s[itex]^{2}[/itex]


    (Time in Air)
    h= Vyt + 1/2gt[itex]^{2}[/itex]
    h= Vsin11t -4.9t[itex]^{2}[/itex]


    (Distance)
    d= Vxt
    10= Vcos11t


    Am I thinking this through right? Or am I way off??
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 2, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi APSurvivor, welcome to Physics Forums.

    You're close to the right track, but not quite there yet.

    You've got expressions for the vertical and horizontal velocities in terms of the (unknown) initial speed, v. That's good.

    When you write the two expressions for the trajectory components, vertical and horizontal, in terms of these velocities and times you got your two expressions:

    [itex] d = v_x t [/itex]
    [itex] h = v_y t - \frac{1}{2} g t^2 [/itex]

    Now, which expression would be simplest to solve for the time to cross the canyon?

    When the person reaches the other side of the canyon, what will be his height h with respect to the launch point?
     
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