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Throwing a ball from atop a hill

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data
    A boy stands at the peak of a hill which slopes downward uniformly at angle ##\phi##. At what angle ##\theta## from the horizontal should he throw a rock so that it has the greatest range?

    2. Relevant equations
    ##\mathbf{s} = \mathbf{v}_0 t + \frac{1}{2} \mathbf{a} t^2##

    ##\mathbf{v} = \mathbf{v}_0 + \mathbf{a} t##

    3. The attempt at a solution
    First I tilted the reference frame so the sloping hill is horizontal and the initial velocity is at an angle ##\theta + \phi## from the ground. This makes the acceleration ##\mathbf{a} = g \sin(\phi) \hat{\imath} - g \cos(\phi) \hat{\jmath}##. So position is given by
    [tex]\mathbf{s} = \left[ v_0 t \cos(\theta + \phi) + \frac12 g t^2 \sin(\phi) \right] \hat{\imath} + \left[ v_0 t \sin(\theta + \phi) - \frac12 g t^2 \cos(\phi) \right] \hat{\jmath}.[/tex] Taking the derivative with respect to time,
    [tex]\mathbf{v} = \left[ v_0 \cos(\theta + \phi) + g t \sin(\phi) \right] \hat{\imath} + \left[ v_0 \sin(\theta + \phi) - g t \cos(\phi) \right] \hat{\jmath}.[/tex]Setting the vertical component of velocity equal to zero gives half the total flight time:
    [tex]t_{\text{max}} = \frac{v_0 \sin(\theta + \phi)}{g \cos(\theta)}.[/tex] Plugging ##2 t_{\text{max}}## into the horizontal component of ##\mathbf{s}## gives the distance traveled:
    [tex]\begin{align}s_x (2 t_{\text{max}}) &= v_0 \frac{2 v_0 \sin(\theta + \phi)}{g \cos(\phi)} \cos(\theta + \phi) + \frac12 g \sin(\phi) \frac{4 v_0^2 \sin^2(\theta + \phi)}{\cos^2(\phi)} \\
    &= \frac{2 v_0^2}{g \cos(\phi)} [ \sin(\theta + \phi) \cos(\theta + \phi) + \sin^2(\theta + \phi) \tan(\phi) ].\end{align}[/tex]
    To find the optimal ##\theta## for a given ##\phi##, take
    [tex]\begin{align}\frac{\text{d}}{\text{d} \theta} s_x (2 t_{\text{max}}) = 0 &= \cos(\theta + \phi) \cos(\theta + \phi) - \sin(\theta + \phi) \sin(\theta + \phi) + 2 \sin(\theta + \phi) \cos(\theta + \phi) \tan(\phi) \\
    &= \cos^2(\theta + \phi) - \sin^2(\theta + \phi) + \sin(2(\theta + \phi)) \tan(\phi) \\
    &= \cos(2(\theta + \phi)) + \sin(2(\theta + \phi)) \tan(\phi).\end{align}[/tex] A bit of algebra gives
    [tex]\tan(2(\theta + \phi)) = - \cot(\phi),[/tex] which results in an equation for ##\theta##
    [tex]\theta = \frac12 \arctan(-\cot(\phi)) - \phi.[/tex]
    Now, the book's hint was that when ##\phi## is 60º, ##\theta## should be 15º. My equation spits out -75º, which has the slight problem of meaning the boy would be throwing the rock through the hill. So my question is why am I off by 90º? Did I make a mistake in my math somewhere? Does it have to do with rotating my reference frame? :frown:
    Last edited: Apr 17, 2013
  2. jcsd
  3. Apr 17, 2013 #2


    Staff: Mentor

    well -75 degrees seems suspiciously like 15 degrees

    so your book says it slopes down at 60 degrees so should you be using -60 degrees?

    I would try to solve it without the rotation to see if things work out.
    Last edited: Apr 17, 2013
  4. Apr 17, 2013 #3
    It seems to me my equation should be spitting out 15º, not 15º - 90º.
  5. Apr 17, 2013 #4
    Using -60º just gives me +75º. I've played with all the signs; the only answer that (sort of) works out is the one I got.

    And solving the problem without the frame rotation is all kinds of unpleasant.
  6. Apr 17, 2013 #5


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    Not really.
    Flight time t, hor dist x, vert dist y (measuring positive downwards), accn a, initial velocity u at θ above horizontal:
    -ut sin(θ) + at2/2 = y = x tan(ϕ) = ut cos(θ)tan(ϕ)
    at/2u = tan(ϕ)cos(θ) + sin(θ)
    xa/u2 = 2tan(ϕ)cos2(θ) + 2 sin(θ)cos(θ) = tan(ϕ)(cos(2θ)+1) + sin(2θ)
    For max x:
    tan(ϕ)(-2sin(2θ)) + 2cos(2θ) = 0
    tan(ϕ) = cot(2θ)
    ϕ = π/2 - 2θ
  7. Apr 17, 2013 #6
    Well, it was the way I was trying to do it, but if anything I've learned that I tend to overcomplicate things.

    Anyway, I figured out the problem was my rotated frame. I plugged the ##2 t_{\text{max}}## I got above into the unrotated horizontal component of the distance equation and got the same equation haruspex did. Though I still don't know why rotating my frame an arbitrary angle offset my answer by exactly 90º.
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