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Homework Help: Max Rage of a Projectile Uphill

  1. Sep 30, 2004 #1
    Help: Max Rage of a Projectile Uphill

    I am working on some physics on a site I found and was wondering if I could tap into some of the wisdom that seems embeded in the forums.

    The problem I am working on is problem three from
    http://www.math.rutgers.edu/~costin/291/w3.pdf [Broken]

    All I am interested in doing is show how to optimize the uphill range.

    I have been doing some inital attacking but don't seem to be getting anywhere. I know the answer from that site [tex] tan 2 \alpha = -cot \theta [/tex]

    Here is some of the stuff that I have been starting out doing
    [tex] y/x=tan\theta [/tex]
    [tex] y= x tan \theta [/tex]
    [tex] R/x= sec \theta [/tex]
    [tex] R = (Vot+1/2at^2) sec \theta [/tex]
    [tex] \frac {v^2-V_o^2}{{2*9.8}}=x [/tex]

    I am not very sure on how many of these are true because I haven't had time fully to reexamine them and I was trying to force things together that probably need to be placed. I will continue to work on this problem as I just started but If anyone can sucessfully show me how to come up with
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Sep 30, 2004 #2


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    Try to make a formula for the distance uphill, then apply calculus to get the maximiun value for the alpha angle. Consider intersection....

    [tex] xtan\theta = tan\alpha x - \frac{g}{2v_{o}^{2}cos^{2}\alpha}x^{2} [/tex]

    [tex] 0 = (tan\alpha - tan\theta)x - \frac{g}{2v_{o}^{2}cos^{2}\alpha}x^{2} [/tex]

    [tex] 0 = x((tan\alpha - tan\theta) - \frac{g}{2v_{o}^{2}cos^{2}\alpha}x) [/tex]

    so one x=0 and the other, you know....

    after finding both X and Y of intersection, a simple apply of the distance formula with 0,0 and X,Y can get you d, after that you could find the max value for alpha using differential calculus.
    Last edited: Sep 30, 2004
  4. Sep 30, 2004 #3
    Still a little lost... I am will look to see if there is any other help when I wake up...;
    for example I am just taking my first calc course Calc BC how would us use calc to ge the max value for the alpha angle...

    ty for your response Cyclovenom
  5. Sep 30, 2004 #4


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    By the way, i edited with some of the work.
  6. Oct 1, 2004 #5
    There must be a way to do it with out differential calculus... the physics course I am taking is made to have no math beyond deravatives and integration in calculus. I was wondering if someone could show me how to derrive

    [tex] tan 2 \alpha = -cot \theta [/tex]

    from the example
  7. Oct 1, 2004 #6


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    First, ignore the slope. A projectile (even a "raging" projectile!) fired with initial speed v0 and angle to the horizontal α has trajectory given by
    x= v0cosα t and y= -(g/2)t2+ v0sinα t.

    If the ground were level, "hitting the ground" would correspond to y= 0

    Here, the "ground" is given by y= x tanθ so the projectile will hit the ground when tan&theta(v0cosα t)= -(g/2)t2+ v0sinα t

    You could solve that quadratic equation for t in terms of α, which is fixed, and θ and then find the θ that maximizes x. I'n sure HOW you would do that without using the derivative!
  8. Oct 2, 2004 #7
    Method One:
    In this method please note that in the picture of the problem
    [tex] \alpha is replaced by \beta [/tex]
    [tex] \theta is replaced by \alpha [/tex]


    [tex]x=Vocos\beta t[/tex]
    [tex]Vosin\Beta t = 1/2gt^2=xtan\alpha=Vocos/beta t tan \alpha [/tex]
    [tex] t^2-\frac2{2}{{g}}(Vosin\beta-Vocos\beta tan\alpha)=0 [/tex]
    [tex] t = (t-\frac2{2}{{g}}(Vosin\beta-Vocos\beta tan\alpha))=0 [/tex]
    [tex] t=0 or t= \frac{2Vo}{{g}}(sin\beta-cos\betatan\alpha[/tex]
    [tex] sec\alpha=\frac{R}{{x}} [/tex]
    [tex] R=xsec\alpha [/tex]
    [tex] =Vocos\beta t sec\alpha [/tex]
    [tex] = Vocos\beta sec\alpha t [/tex]
    [tex] =Vocos\beta sec\alpha (\frac{2vo}{{g}}(sin\beta-cos\beta tan\alpha)) [/tex]
    [tex] =\frac{2vo^2}{{g}}\frac{sin\beta cos\beta}{{cos\alpha}}-\frac{cos^2\beta}{{cos\alpha}}\frac{sin\alpha}{{cos\alpha}} [/tex]
    [tex] \frac{2Vo^2}{{g}} (\frac{sin\beta cos\beta cos\alpha - cos^2\beta sin\alpha}{{cos^2\alpha}} [/tex]
    [tex] \frac{2Vo^2}{{g}}(\frac{sin\beta cos\alpha - cos\beta sin \alpha }{{cos^2\alpha}})cos\beta [/tex]
    [tex] {2Vo^2}{{g}} \frac{{sin(alpha-\beta)cos\beta}{{cos^2\alpha}} [/tex]
    [tex] R= \frac{2Vo^2}{{g}}(\frac{sin(\alpha - \beta)cos\beta}{{cos^2\alpha}}) [/tex]
    [tex] R= \frac{2Vo^2}{{g}}(\frac{sin(2\beta-\alpha)-sin\alpha}{{cos^2\alpha}} [/tex]
    since everything is constant minus [tex]sin(2/beta-/alpha) [/tex] you must maximize it[/tex]

    [tex] 2\beta = \frac{\pi}{{2}}+\alpha [/tex]
    [tex] \beta = \frac{\pi}{{4}}+\frac{\alpha}{{2}} [/tex]
    Last edited: Oct 2, 2004
  9. Oct 2, 2004 #8
    Does this seem alright
  10. Oct 2, 2004 #9


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    Tom, will you please fix the Latex? oh and Yes the answer is correct.
    Last edited: Oct 3, 2004
  11. Oct 4, 2004 #10
    I can't edit the post any more otherwise I would
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