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Horizontal Range (with incline)

  1. May 26, 2008 #1
    [SOLVED] Horizontal Range (with incline)

    1. The problem statement, all variables and given/known data

    A ball is thrown up an (v0) incline. The ball is thrown at an angle phi from the incline and the incline has an angle theta with the ground.
    1) Show the ball lands a distance [tex]R = 2{v_0}^2 sin\theta cos(\theta + \phi ) / (gcos{\phi }^2[/tex]

    2) Show that for a given v0 and phi, the maximum possible range up the incline is [tex]R_{max} = {v_0}^2/(g(1+sin\phi))[/tex]

    3. The attempt at a solution


    I was able to do 1) fairly easily, however, I'm having a bit of trouble with 2).
    I was thinking that since the max range on a horizontal plane is 45 degrees, I could just shift theta by 45 degrees and say [tex]\theta = 45 + \phi [/tex] unfortunately, that doesn't seem to lead me to the correct answer.

    I was then thinking of taking the derivative of R and seeing if I could find the maximum of it that way.

    [tex]dR/d\theta = cos\theta cos(\theta + \phi) - sin \theta sin( \theta + \phi ) = 0[/tex]
    Unfortunately this seems to lead me nowhere (don't know how to solve for theta)

    Does anyone know how I should go about finding theta? I was hoping that it was going to be 45 degrees, but I don't think that it is, and now I don't know where to go.
     
  2. jcsd
  3. May 27, 2008 #2

    alphysicist

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    Homework Helper

    Hi jesuslovesu,

    Could the statement of the problem be wrong here? It states that the angle phi is the launch angle of the ball relative to the incline, and theta is the angle of the incline itself (relative to the ground). Were those supposed to be switched?

    If those definitions were true, then for level ground, theta=0, which means that your range formula in part 1 gives zero (since it has a sin(theta) in the numerator).
     
  4. May 27, 2008 #3
    Oh my bad, sorry yes, theta is the angle of the ball's velocity and phi is the angle of the incline.
     
  5. May 27, 2008 #4
    [tex]s=Rcos\phi[/tex]
    [tex]v_{YO}=v_{0}sin(\phi+\theta)[/tex]
    [tex]v_{XO}=v_{0}cos(\phi+\theta)[/tex]
    [tex]h=v_{YO}t-\frac{gt^{2}}{2}[/tex]
    [tex]s=v_{XO}t[/tex]
    so
    [tex]t=\frac{Rcos\phi}{v_{0}cos(\phi+\theta)}[/tex]
    so
    [tex]\frac{gR^{2}cos^{2}\phi}{2v_{0}^{2}cos^{2}(\phi+\theta)}+Rsin\phi-\frac{Rsin(\phi+\theta)cos\phi}{cos(\phi+\theta)}=0[/tex]
    so
    [tex]R=0[/tex] or [tex]R=\frac{2v_{0}^{2}cos(\phi+\theta)sin\theta}{cos^{2}\phi}[/tex]
     
  6. May 27, 2008 #5
    [tex]\frac{dR}{d\theta}=cos\theta{cos(\phi+\theta)}-sin\theta{sin(\phi+\theta)}=0[/tex]
    but
    [tex]cos(\phi+\theta)=cos\theta{cos\phi}-sin\theta{sin\phi}[/tex]
    [tex]sin(\phi+\theta)=sin\phi{cos\theta}+cos\phi{sin\theta}[/tex]
    to equation
    [tex]\frac{dR}{d\theta}=cos\phi(cos^{2}\theta-sin^{2}\theta)-2sin\theta{cos\theta}sin\phi=0[/tex]
    and
    [tex]cos2\theta=cos^{2}\theta-sin^{2}\theta[/tex]
    [tex]sin2\theta=2sin\theta{cos\theta}[/tex]
    so
    [tex]tg2\theta=ctg\phi[/tex]
     
    Last edited: May 27, 2008
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