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Math Help - Projectile Motion Question #2

  1. #1
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    Projectile Motion Question #2

    I'm stuck on how to approach these 2 problems - any hints would be much appreciated!
    1.
    A body is projected form a point O on horizontal ground with velocity of 70m/s. It passes through a point P, which is 45m above the ground and 50m from O horizontally. Find the tangents of the two possible angles of elevation of the projected body. Find also the gradients at P of the two paths.
    2.
    The maximum range of a certain gun on a horizontal plane is R. Find the two possible angles at which the gun can be fired to give a range of 1/2R. If T1 and T2 are corresponding times of flight, find T1:T2.
    Last edited by mr fantastic; February 13th 2011 at 03:18 PM. Reason: Title.
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  2. #2
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    Hello, ksinger!

    I'll get you started on #1 . . .


    \text{1. A body is projected form a point }O\text{ on horizontal ground}
    \text{with a velocity of 70m/s.\; It passes through a point }P\text{ which is}
    \text{45m above the ground and 50m from }O\text{ horizontally.}

    \text{(a) Find the tangents of the two possible angles of elevation}
    . . . . \text{of the projected body.}

    We are expected to know these two parametric equations for projectile motion.

    . . \begin{array}{c}x \;=\;(70\cos\theta)t \qquad\quad  \\ y \;=\;(70\sin\theta)t - 4.9t^2 \end{array} \quad\text{where }\theta\text{ is the angle of elevation}

    If \,O is at the origin, then point \,P is (50,45).


    We have: . x = 50\quad\Rightarrow\quad (70\cos\theta)t \:=\:50 \quad\Rightarrow\quad t \:=\:\dfrac{5}{7\cos\theta} .[1]
    And: . y = 45 \quad\Rightarrow\quad (70\sin\theta)t - 4.9t^2 \:=\:45 .[2]


    Substitute [1] into [2]:

    . . 70\sin\theta\left(\dfrac{5}{7\cos\theta}\right) - 4.9\left(\dfrac{25}{49\cos^2\!\theta}\right) \;=\;45

    . . 50\tan\theta - 2.5\sec^2\!\theta \;=\;45 \quad\Rightarrow\quad 50\tan\theta - 2.5(\tan^2\!\theta + 1) \;=\;45

    . . 50\tan\theta - 2.5\tan^2\!\theta - 2.5 \;=\;45 \quad\Rightarrow\quad 2.5\tan^2\!\theta - 50\tan\theta + 47.5\;=\;0


    Multiply by 0.4: . \tan^2\!\theta - 20\tan\theta + 19 \;=\;0

    Factor: . (\tan\theta - 1)(\tan\theta - 19) \;=\;0

    . . \text{Therefore: }\begin{Bmatrix}\tan\theta &=& 1 \\ \tan\theta &=& 19 \end{Bmatrix}

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