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Homework Help: Tossing an item up a slope, calculating it's length from start to landing

  1. Aug 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Right so I have the following scenario: An archer shoots an arrow upward a slope of angle [tex]\alpha[/tex], the angle between the arrows trajectory and the horizontal plane is denoted
    [tex]\theta[/tex]
    The arrows speed in the direction in which it is fired is denoted [tex]V_0[/tex]
    as shown in this crudely drawn picture:

    http://img124.imageshack.us/img124/8735/arrownb4.jpg [Broken]

    My task is to find the formula for the length between the arrows initial position to the position where it lands, for all [tex]V_0[/tex], [tex]\theta[/tex] and [tex]\alpha[/tex]
    2. Relevant equations
    I know the answer to the question is supposed to be:
    [tex]\frac{2V^{2}_{o}cos^2\theta}{g*cos\alpha} * (tan\theta - tan\alpha)[/tex]


    3. The attempt at a solution
    Well I started off by thinking that the arrows position measured by time for x and y coordinates respectively are:

    [tex] x(t) = V_{o} * cos\theta * t [/tex]

    [tex]y(t) = V_{o} * sin\theta * t - \frac{1}{2} * g * t^2[/tex]

    I went on to show that as a function of x, y can be written:
    [tex]y(x) = \frac{V_0y}{V_0x}x - \frac{g}{2v^{2}_{ox}}x^2[/tex]
    Since I know that on this items trajectory it has to stop where it intersects with the line [tex] y(x) = \alpha x[/tex]

    I put:
    [tex]\alpha x = \frac{V_oy}{V_0x}x - \frac{g}{2v^{2}_{ox}}x^2[/tex]

    Now, replacing x with [tex]V_o cos\theta t[/tex] and [tex]V_oy[/tex] and [tex]V_ox[/tex] with their respective formulas of t and sin/cos [tex]\theta[/tex] I eventually wound up with the following expression of t

    [tex] t = \frac{(tan\theta - \alpha)2cos\theta}{g}[/tex]

    and further more figured that since the length from the starting position to the position where the arrow lands is [tex]\sqrt{x^2 + y^2}[/tex] and I now have an expression for t that only uses constants, and I also know formulas for the position of x and for the position of y, I could replace the t in these formulas and put them in the root above to gain an expression for the length L. Such as this:

    [tex] x(t) = V_{o} * cos\theta * t [/tex]
    [tex]y(t) = V_{o} * sin\theta * t - \frac{1}{2} * g * t^2[/tex]
    [tex] t = \frac{(tan\theta - \alpha)2cos\theta}{g}[/tex]
    [tex] L = \sqrt{x^2 + y^2}[/tex]
    [tex] L = (\sqrt{V_{o} * cos\theta * \frac{(tan\theta - \alpha)2cos\theta}{g})^2 + (V_{o} * sin\theta * \frac{(tan\theta - \alpha)2cos\theta}{g} - \frac{1}{2} * g * (\frac{(tan\theta - \alpha)2cos\theta}{g})^2 ) ^2}} [/tex]

    However as most of you can probably observe this is a right mess and most likely won't lead anywhere at all, and i'm willing to bet my left leg there's something quite easy I should be able to observe in order to solve the problme, I just can't seem to put my finger on it. Anyone willing to give me a nudge in the right direction would be greatly appreciated.

    So anyone have any better idea on how to solve this?
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Aug 21, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Molecular! :smile:

    Yes, your proof looks fine. :smile:

    (though you should be using tanα instead of α, and you could have left out the y(x) formula, and just put y = x tanα in the y(t) and x(t) formulas :wink:)

    To get L, I'd recommend simplifying by defining µ (say) = 2cosθ (tanθ - tanα)/g …

    the formula should then look quite neat. :smile:
     
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