1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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


    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):

    cos(2a)=1-2sin^2 (a), so


    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])
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook