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Math Help - calculus, finding velecity of shadow

  1. #1
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    calculus, finding velecity of shadow

    here is the question:

    a light source is dropped from a \frac{1}{2}gT^2 height tower top (t=0).
    from a point which is a distant from the tower base, a stone is thrown with velocity of gT vertically at the same time.
    find the velocity of the stone's shadow when t=T/4

    i can find velocity of the light source of the light source and stone at a t=t.
    but can't find the velocity of the shadow using them.
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  2. #2
    MHF Contributor
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    Displacement and velocity

    Hello stud_02
    Quote Originally Posted by stud_02 View Post
    here is the question:

    a light source is dropped from a \frac{1}{2}gT^2 height tower top (t=0).
    from a point which is a distant from the tower base, a stone is thrown with velocity of gT vertically at the same time.
    find the velocity of the stone's shadow when t=T/4

    i can find velocity of the light source of the light source and stone at a t=t.
    but can't find the velocity of the shadow using them.
    I assume that the shadow is cast on the ground, which is horizontal.

    Then the way to do this is as follows:

    • Use the equations of constant acceleration to find an expression for the height, y_L, above the ground of the light source at time t.


    • Find an expression for the height, y_S, above the ground of the stone at time t.


    • Draw a straight line from the light source to the stone, and produce it until it meets the ground. The point where it meets the ground is the position of the shadow.


    • Use similar triangles to find the distance, x, of the shadow from the foot of the tower, in terms of y_L, y_S, and hence in terms of t.


    • Differentiate x with respect to t to get an expression for \dot{x}, the velocity of the shadow in terms of t.


    • Put t = \tfrac14T to find the answer.

    I'll give you one or two steps:

    (1) y_L = \tfrac12gT^2 - \tfrac12gt^2

    (2) y_S = ... ?

    (3) x = \frac{ay_L}{y_L-y_S}=... ?

    (4) \dot{x} = ... ?

    If I haven't made any mistakes with the algebra, I think the answer is \frac{13a}{2T}.

    Grandad
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  3. #3
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    Hello, stud_02!

    Wow! . . . What a long, intricate problem!


    A light source is dropped from a \tfrac{1}{2}gT^2 foot tower top (t=0).

    From a point which is a distant from the tower base,
    a stone is thrown with velocity of gT vertically at the same time.

    Find the velocity of the stone's shadow when t=\tfrac{T}{4}
    Code:
                                  A
                                  *   -   -
                                  |   :   :
                                  |   :  16gtē
                                  |   :   :
                                  * B :   -
                               *  |   :   :
                            *     | ―gTē  :
                      D  *        |   :   :
                      o         y |   :  16gTē-16gtē
                   *  |           |   :   :
                *     |gTt        |   :   :
             *        |           |   :   :
          o-----------*-----------*   -   -
          S    x-a    E     a     C
          : - - - - - x - - - - - :

    The tower is: AC = \tfrac{1}{2}gT^2

    In t seconds, the light falls: AB = 16gt^2 feet.

    Hence: . y \:=\:BC \:=\:16gT^2 - 16gt^2\;\;{\color{blue}[1]}


    The stone is thrown upward from E\!:\;DE = gTt feet.

    We have: EC = a

    The shadow of the stone is at S\!:\;\text{ let }x = SC.
    . . Then: SE \:=\: x-a


    We have similar right triangles: . \Delta BCS \sim \Delta DES

    So we have: . \frac{x}{y} \:=\:\frac{x-a}{gTt} \quad\Rightarrow\quad x \:=\:a\cdot\frac{y}{y-gTt}

    Substitute [1]: . x \;=\;a\cdot\frac{\frac{1}{2}g(T\:\!^2-t^2)}{\frac{1}{2}g(T\:\!^2-t^2) - gTt} \;=\;\frac{\frac{1}{2}g(T\:\!^2-t^2)}{\frac{1}{2}g(T\:\!^2-t^2 - 2Tt)}

    . . and we have: . x \;=\;a\cdot\frac{T\:\!^2-t^2}{T\:\!^2-t^2-2Tt}


    Differentiate: . \frac{dx}{dt} \;=\;a\cdot\frac{(T\:\!^2 - t^2-2Tt)(-2t) - (T\:\!^2-t^2)(-2t-2T)}{(T\:\!^2-t^2-2Tt)^2}

    . . which simplifies to: . \frac{dx}{dt} \;=\;\frac{2aT(T\:\!^2 + t^2)}{(T\:\!^2 + t^2 - 2Tt)^2}


    Now plug in t = \frac{T}{4}



    But check my work . . .please!
    .
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  4. #4
    MHF Contributor Calculus26's Avatar
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    Animation

    Excellent problem.

    To all concerned-- I'll be adding an animation on my website
    Calculus Animations

    later today demonstrating this problem.

    Just go to the related rates page and look for falling light source.

    By the way grandad your algebra is perfect.



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