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Math Help - Projectiles question

  1. #1
    Junior Member silvercats's Avatar
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    Projectiles question

    Object is projected from a velocity of 18 m/s and angle of 60 degrees .
    take g=10

    what is the flying time , Horizontal distance and the maximum height of the object ?
    this is what I did which gave the wrong answers
    v=u+at
    0=v sin Θ-gt/2
    t=2v*sin Θ/g
    t=2*18 sin 60/10
    t=1.8*√3/2
    t=1.56

    -->
    s=18*cos 60*1.56 =14m

    (up)
    s=18*√3/2 - 5(1.56)
    =3.42m

    Those are not answers .help
    thanks



    book says 0.77s << I think time is wrong in the book:
    Horizontal distance 28.02m (i got 14)
    maximum height 12.15m (i got 3.42)
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  2. #2
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    Hello, silvercats!

    Object is projected with a velocity of 18 m/s and angle of 60 degrees. .Take g=10

    What is the (a) flying time, (b) horizontal distance
    and (c) the maximum height of the object?

    The equations for projectiles are: . \begin{Bmatrix}x \;=\; (v_o\cos\theta)t \\ y \;=\; h_o + (v_o\sin\theta)t - \frac{1}{2}gt^2 \end{Bmatrix}

    . . where: . \begin{Bmatrix}h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \\ \theta &=& \text{angle of elevation} \end{Bmatrix}


    We have: . h_o = 0,\;\;v_o = 18,\;\;\theta = 60^o,\;\;g=10

    Hence: . x \;=\;(18\cos60^o)t \quad\Rightarrow\quad x \;=\;9t
    . . . . . . y \;=\;(18\sin60^o)t - \tfrac{1}{2}(10)t^2 \quad\Rightarrow\quad y \;=\;9\sqrt{3}t - 5t^2


    (a) Flying time: When does the object strike the ground?
    . . .When y = 0.

    . . 9\sqrt{3}t - 5t^2 \:=\:0 \quad\Rightarrow\quad t(9\sqrt{3} - 5t) \:=\:0 \quad\Rightarrow\quad t \:=\:0,\;\tfrac{9\sqrt{3}}{5}

    . . \text{The object flies for }\frac{9\sqrt{3}}{5}\text{ seconds.}


    (b) Horizontal distance.
    . . . In \tfrac{9\sqrt{3}}{5}\text{ seconds, the object travels}
    . . . . . x \:=\:9\left(\tfrac{9\sqrt{3}}{5}\right) \:=\:\frac{81\sqrt{3}}{5}\text{ meters horizontally.}


    (c) Maximum height
    . . .The object reaches maximum height in \tfrac{9\sqrt{3}}{10}\text{ seconds.}
    . . .(Half the flying time.)

    \text{The maximum height is:}
    . . y \;=\;9\sqrt{3}\left(\tfrac{9\sqrt{3}}{10}\right) - 5\left(\tfrac{9\sqrt{3}}{10}\right)^2 \;=\;\frac{243}{20} \:=\:12.15\text{ meters.}


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  3. #3
    Junior Member silvercats's Avatar
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    Quote Originally Posted by Soroban View Post
    (a) Flying time: When does the object strike the ground?
    . . .When y = 0.

    . . 9\sqrt{3}t - 5t^2 \:=\:0 \quad\Rightarrow\quad t(9\sqrt{3} - 5t) \:=\:0 \quad\Rightarrow\quad t \:=\:0,\;\tfrac{9\sqrt{3}}{5}

    . . \text{The object flies for }\frac{9\sqrt{3}}{5}\text{ seconds.}
    But can't we take like this?
    9√3-5t=0
    5t=9√3
    t=(9√3)/5
    t= sqrt{(9√3)/5} ?
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