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Homework Help: Derive optimal anlge for maximum range (Projectile Motion)

  1. May 3, 2012 #1
    1. The problem statement, all variables and given/known data

    For an assignment I need to derive the optimal angle for maximum range (derive Equation 2 below). I know how to derive equation 1 but I need to derive the second equation so I can substitute h = 0, and h = 1, into it, to show the optimal angle for maximum range for bi-level and uni-level.

    I have a book that explains it but I don't understand it at all. Any help is really appreciated.

    Thank you!

    2. Relevant equations
    Equation 1:
    Rmax = u^2 / g √(1 + 2gh/u^2)

    Equation 2:
    tan θ = 1 / √ (1 + 2gh / u^2)

    u = initial velocity
    g = grav. accel.
    h = height

    3. The attempt at a solution
    One method I have tried is finding the derivative of Rmax, setting it to zero and solving for theta (to find the minimum) which does give me 45 degrees but I cannot apply this to bi-level projection.
  2. jcsd
  3. May 4, 2012 #2
    The way you've written the fractions is somewhat ambiguous. It's hard to say whether you meant that [itex]R_{\text{max}}=\frac{u^2}{g}\frac{1}{\sqrt{1+\frac{2gh}{u^2}}}[/itex] or [itex]R_{\text{max}}=\frac{u^2}{g}\sqrt{1+\frac{2gh}{u^2}}[/itex] (especially since the term in the radical is dimensionless), but I've realized that it's the second one - the moral of the story is to use parentheses :smile:.

    In this case, one method is to use the standard [itex]x(t)[/itex] and [itex]y(t)[/itex] to find [itex]y(x)[/itex] and sub in [itex]y(R)=0[/itex] to get an equation that defines [itex]R[/itex] implicitly as a function of [itex]\theta[/itex]. Knowing this, you can find [itex]\tan(\theta_{\text{max}})[/itex] as a function of [itex]R_{\text{max}}[/itex], which you said you already found.
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