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Math Help - If you shoot an arrow, how do you find out how far it will go?

  1. #1
    av8
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    If you shoot an arrow, how do you find out how far it will go?

    Theres the question:

    A person shoots an arrow which leaves the bow at 4ft above the ground with an initial velocity of 88ft second at an elevation angle of 48*.

    Will the arrow go over a 40 foot wall that is 200' away?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Um... I will use kinematics to solve this.

    The vertical component of the arrow is

    88 \sin 48 = 65.4 ft/s

    The horizontal component of the arrow is 88 \cos 48 = 58.9 ft/s

    So, the time it takes to cover 200 ft is 200/58.9 = 3.40 s.

    Now, use the formula s = ut + \frac12 at^2

    with t = 3.40,
    u = 65.4 ft/s
    a = -9.8 m/s^2 (you will have to convert to ft/s^2)

    and solve for s.

    If s < (40 - 4), then the arrow won't pass. Otherwise, it will.

    What did you get?
    Last edited by Unknown008; December 9th 2010 at 09:49 AM. Reason: Forgot the 4ft throw point.
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  3. #3
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    Hello, av8!

    \text{A person shoots an arrow which leaves the bow 4 ft above the ground}
    \text{with an initial velocity of 88 ft/sec at an elevation angle of 48}^o.

    \text{Will the arrow go over a 40-foot wall that is 200 feet away?}

    It certainly helps if you know the Trajectory Equations.

    . . . . . x \;=\;v_o\cos\theta\, t

    . . . . . y \;=\;h_o + v_o\sin\theta\,t - 16t^2

    . . where: . \begin{Bmatrix}<br />
h_o &=& \text{initial height} \\<br />
 v_o &=& \text{initial velocity} \\<br />
\theta &=& \text{initial angle} \end{Bmatrix}



    We are given: . h_o = 4,\;v_o = 88,\;\theta = 48^o


    Our equations are:

    . . x \;=\;(88\cos48^o)t

    . . y \;=\;4 + (88\sin48^o)t - 16t^2


    The question is: When x = 200, is y \,\ge\,40 ?


    When x = 200, we have:
    . . (88\cos48^o)t \:=\:200 \quad\Rightarrow\quad t \:=\:\dfrac{200}{88\cos48^o} \:=\:3.396537613

    It takes about 3.4 seconds to travel 200 feet horizontally.


    When t = 3.4, we have:

    . . y \;=\;4 + (88\sin48^o)(3.4) - 16(3.4^2) \;\approx\;41.4\text{ feet}


    Yes, the arrow will clear the wall.

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