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Projectile thrown up an incline, max distance?

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown with initial speed [tex] V [/tex] up an inclined plane. The plane is inclined at an angle [tex]\phi[/tex] above the horizontal, and the ball's velocity is at an angle [tex]\theta[/tex] above the plane.

    Show that the ball lands a distance [tex]R = \frac{ 2V^2 \sin{\theta} \cos{ ( \theta + \phi} ) } { g \cos^2{\phi} }[/tex] from its launch point. Show that for a given [tex]V[/tex] and
    [tex]\phi[/tex], the maximum possible range up the inclined plane is
    [tex] R_{max} = \frac{ V^2 }{ g (1 + \sin{\phi} )} [/tex]


    2. Relevant equations

    F = ma

    3. The attempt at a solution

    I calculated the distance traveled up the incline fine. However, I'm having trouble proving the second part. I'm guessing I'm supposed to maximize R with respect to theta, so from the equation above we have:

    [tex]\frac{d} {d \theta} \sin{\theta} \cos{(\theta + \phi)} = 0[/tex]

    [tex]\cos{( 2 \theta + \phi )} = 0[/tex]

    [tex]\theta = \frac{n \pi}{4} - \frac{\phi}{2} [/tex], with n = odd integer. Now plug this back into the original equation for R and I get

    [tex]R_{max} = \frac{ 4V^2 \tan{\phi} }{g} \cos{ 2\phi} [/tex]

    I think I'm approaching this problem wrong. Can anyone give me a simpler way?
     
  2. jcsd
  3. Jun 16, 2009 #2
  4. Jun 16, 2009 #3
    if you look at the last post in the thread you gave me, I already got that result. It's just that the R_max I get does not match the R max that's given in the initial problem.
     
  5. Jun 16, 2009 #4
    Ops my bad.
     
  6. Jun 16, 2009 #5

    ideasrule

    User Avatar
    Homework Helper

    First, n must be 1; if it's 3 or higher, than n*pi/4 - Φ/2 would be greater than 90 degrees, and that's impossible given the geometry of the situation. So we know that the optimum angle is pi/4-Φ/2. Substituting it into the equation for R gives:

    R=sin(pi/4-Φ/2)cos(pi/4+Φ/2)/cos^2 (Φ)

    I've ignored the constants. Since cos(a)=sin(pi/2-a):

    R=sin(pi/4-Φ/2)sin(pi/4-Φ/2)
    cos(2a)=1-2sin^2 (a), so

    R=(1-cos(pi/2-Φ))/2
    =(1-sin(Φ))/2

    You seem to have decent math skills, so getting from here to the answer should be easy.
     
  7. Jun 16, 2009 #6
    you have got [tex]\theta[/tex] = [tex]\pi/4[/tex] - [tex]\varphi/2[/tex]
    so 2sin[tex]\theta[/tex] cos([tex]\theta[/tex] + [tex]\varphi[/tex]) = 2sin([tex]\pi/4[/tex] - [tex]\varphi/2[/tex]) sin([tex]\pi/4[/tex] - [tex]\varphi/2[/tex]) =
    2sin2([tex]\pi/4[/tex] - [tex]\varphi/2[/tex]) = [cos([tex]\varphi/2[/tex]) - sin([tex]\varphi/2[/tex])]2

    [cos([tex]\varphi/2[/tex]) - sin([tex]\varphi/2[/tex])]2/cos2[tex]\varphi[/tex]
    = {[cos([tex]\varphi/2[/tex]) - sin([tex]\varphi/2[/tex])]/cos[tex]\varphi[/tex]}2
    = {[cos([tex]\varphi/2[/tex]) - sin([tex]\varphi/2[/tex])]/[cos2([tex]\varphi/2[/tex]) - sin2([tex]\varphi/2[/tex])]}2
    = {1/[cos([tex]\varphi/2[/tex]) + sin([tex]\varphi/2[/tex])]}2
    = 1/(1 + sin[tex]\varphi[/tex])
     
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