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Math Help - Motion in Three-Space

  1. #1
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    Motion in Three-Space

    A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Neglect air resistance and use as the acceleration of gravity.
    Answer in radians.

    Thanks!
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  2. #2
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    Quote Originally Posted by jffyx View Post
    A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Neglect air resistance and use as the acceleration of gravity.
    Answer in radians.

    Thanks!
    It will depend on what height you want to hit the object at. Or is it to be assumed that the object is at ground level ....?
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    Quote Originally Posted by mr fantastic View Post
    It will depend on what height you want to hit the object at. Or is it to be assumed that the object is at ground level ....?
    Oh. It is to be assumed that the object is at ground level.
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    Quote Originally Posted by jffyx View Post
    Oh. It is to be assumed that the object is at ground level.
    Substitute the data into equation (7) here: Projectile Motion

    Now solve this equation for \theta.
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  5. #5
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    Hello, jffyx!

    A gun has a muzzle speed of 90 meters per second.
    What angle of elevation should be used to hit an object 150 meters away?
    The equations for projectile motion are: . \begin{array}{ccc}x \;=\; (v_o\cos\theta)t \\ y \;=\; h_o + (v_o\sin\theta)t - 4.9t^2 \end{array}
    . . where: . \begin{Bmatrix}h_o &=& \text{initial height} \\ v_o &=& \text{initial speed} \\ \theta &=& \text{angle of elevation} \end{Bmatrix}


    We are given: . h_o = 0,\;v_o=90

    The equations are: . \begin{array}{cc}x \;=\;(90\cos\theta)t & {\color{blue}[1]} \\ y \;=\;(90\sin\theta)t - 4.9t^2 & {\color{blue}[2]} \end{array}

    Since x = 150, [1] becomes: . (90\cos\theta)t \:=\:150 \quad\Rightarrow\quad t \:=\:\frac{5}{3\cos\theta}

    Substitute into y = 0\!:\;\;(90\sin\theta)\left(\frac{5}{3\cos\theta}\  right) - 4.9\left(\frac{5}{3\cos\theta}\right)^2 \:=\:0

    which simplifies to: . 1350\tan\theta - 122.5\sec^2\!\theta \:=\:0

    . . 1350\tan\theta - 122.5(\tan^2\!\theta + 1) \:=\:0 \quad\Rightarrow\quad 122.5\tan^2\!\theta - 1350\tan\theta + 122.5 \:=\:0

    Quadratic Formula: . \tan\theta \;=\;\frac{1350 \pm\sqrt{1,\!762,\!475}}{245} \;=\;\begin{Bmatrix}0.091500952 \\ 10.92890771 \end{Bmatrix}


    . . \tan\theta \,=\,0.091500952\quad\Rightarrow\quad \theta \:\approx\:5.228^o

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  6. #6
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    Note how Soroban got two solutions. You could also fire the gun at about 84.77 degrees and hit the target.
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