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Math Help - Projectile motion question

  1. #1
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    Projectile motion question

    Hi. I'm having trouble solving this problem:

    "A prjoectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection?"

    I tried using the formulas for x and y displacement as so:

    x-displacement = vcosθt
    y-displacement = vsinθt - 1/2gt^2

    therefore

    vcosθt = 3(vsinθt - 1/4gt^2) where time = 1/2t because this is when the projectile is at its highest point

    But I can't get rid of the t's in the equation so I think I'm going about this the wrong way. Could somebody point me in the right direction? Thanks
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  2. #2
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    The horizontal range can be determined by finding the times at which the vertical dispacement is zero.

    v \sin(\theta) \cdot t - \frac{1}{2} \cdot g \cdot t^{2} = 0

    This has two solutions:

    t = 0 -- That's when you fired the projectile.

    t_{f} = \frac{2 \cdot v \cdot \sin(\theta)}{g} -- This is when it lands.

    t_{m} = \frac{v \cdot \sin(\theta)}{g} -- This is the middle of the trip. Perhaps this is the moment of greatest height?

    Now, what was the question?

    Horizontal Range: v \cdot \cos(\theta) \cdot t_{f}

    3x Max Height: 3 \cdot \left(v \sin(\theta) \cdot t_{m} - \frac{1}{2} \cdot g \cdot t_{m}^{2}\right)

    You're almost done. What's next?
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  3. #3
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    Cool thanks TKHunny now I've equated those as so:

    vcosθ . (2vsinθ)/g = 3[vsinθ . (vsinθ)/g - 1/2g . (vsinθ)/g]

    and managed to solve that down to

    4vcosθ = 6vsinθ - 3g

    but now I'm stuck because I've still got more than one variable. Do I need to find v in terms of θ or something? Sorry about the unreadable equations as well I haven't figured out how to use latex yet
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  4. #4
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    My mistake, I forgot to square the t...all fixed now. I eventually got it down to

    tanθ = 4/3
    θ = 53.13

    Which was the correct answer! Thanks for your help TK
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