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Math Help - [SOLVED] Projectiles....

  1. #1
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    Unhappy [SOLVED] Projectiles....

    A stone is projected from a point O on a cliff with a speed of 20 m/s at an angle of elevation of 30(degrees) . T seconds later the angle of depression of the stone from O is 45(degrees). Find the value of T .

    i encountered this problem while solving M2 past exam quetions and i couldnt solve it... ... to make things worse no one in my class knows how to solve this problem . Pls hlp me cuz xam is in a few days. i will be very grateful...
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    I am including a diagram. I hope I am interpreting correctly.

    We can use the horizontal and vertical component formulas.

    horizontal: x=(20cos(30))t

    vertical: y=(20sin(30))t-\frac{1}{2}(9.8)t^{2}

    When the rock is level with the point it was thrown, then it's the point where y equals 0, we can find x at that time.

    0=20sin(30)t-\frac{1}{2}(9.8)t^{2}

    Solving for t we find t=2.04 seconds to reach that point.

    That means it's x coordinate is 20cos(30)(2.04)=35.35 \;\ m

    The stone is at (35.35,0) at 2.04 seconds.

    Now, we have a triangle with angle 45 and side 35.35

    So, the y component is the same, y=-35.35 feet

    Plug this into the y component and solve for t.

    -35.35=20sin(30)t-\frac{1}{2}(9.8)t^{2}

    And we find t=3.89 seconds

    Perhaps TopsQuark will check me out.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  3. #3
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    Hello, PFX!

    A stone is projected from a point O on a cliff with a speed of 20 m/s
    at an angle of elevation of 30.
    t seconds later the angle of depression of the stone from O is 45.
    Find the value of t.

    The parametric equations for the position of the stone (relative to O) are:

    . . \begin{array}{ccccccc}x &=& (20\cos30^o)t & \Rightarrow & x &=& 10\sqrt{3} t \\<br />
y &=& (20\sin30^o)t - 4.9t^2 & \Rightarrow & y &=& 10t - 4.9t^2 \end{array}

    Consider the position of the stone at time t.
    Code:
             *
           *   *
         O* - - - +A
            \    *:
              \   :
                \ :
                  *B

    Since \angle AOB = 45^o,\:\Delta OAB is an isosceles right triangle.
    . . Hence: .  OA = AB

    This means: . x \,=\,-y\quad\Rightarrow\quad (10\sqrt{3})t \;=\;-\left[10t - 4.9t^2\right]

    . . which simplfies to: . 4.9t^2 - 10(1+\sqrt{3})t \:=\:0

    . . which factors: . t\left[4.9t - 10(1+\sqrt{3})\right] \:=\:0

    . . and has roots: . t = 0,\;\frac{10(1+\sqrt{3})}{4.9}

    Therefore: . t \:\approx\:5.6 seconds.

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    Eater of Worlds
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    My thinking was on line with yours, Soroban. Why is my answer different?.

    I don't see anything wring with my logic. The range of the stone is \frac{400sin(60)}{9.8}=35.33.
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  5. #5
    PFX
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    hi, thanx for ur replies....

    well the correct answer given is T = 5.59 (to 3 s.f.)

    so Soroban's answer is correct

    @ galactus i cant find where u went wrong....
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    Quote Originally Posted by galactus View Post
    I am including a diagram. I hope I am interpreting correctly.

    We can use the horizontal and vertical component formulas.

    horizontal: x=(20cos(30))t

    vertical: y=(20sin(30))t-\frac{1}{2}(9.8)t^{2}

    When the rock is level with the point it was thrown, then it's the point where y equals 0, we can find x at that time.

    0=20sin(30)t-\frac{1}{2}(9.8)t^{2}

    Solving for t we find t=2.04 seconds to reach that point.

    That means it's x coordinate is 20cos(30)(2.04)=35.35 \;\ m

    The stone is at (35.35,0) at 2.04 seconds.

    Now, we have a triangle with angle 45 and side 35.35

    So, the y component is the same, y=-35.35 feet

    Plug this into the y component and solve for t.

    -35.35=20sin(30)t-\frac{1}{2}(9.8)t^{2}

    And we find t=3.89 seconds

    Perhaps TopsQuark will check me out.
    Take a look at your diagram. When y = 0 you have x at 35.35 ft. How can it be that when it drops to -45 degrees? x is always increasing.

    -Dan
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