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Math Help - Forming a quadratic equation/Physics motion quesiton?

  1. #1
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    Forming a quadratic equation/Physics motion quesiton?

    A particle is observed to to pass through a distance d horizontally and d/10 vertically from the point of projection. Show that if T= tan (theta) then T satisfies the equation:

    75T^2 - 40T + 4 =0

    What is confusing me is how to do this without values as I have an equation to form the quadratic equation:

    x tan(theta) - x^2(g/2u^2)sec^2(theta)
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by thermalwarrior View Post
    A particle is observed to to pass through a distance d horizontally and d/10 vertically from the point of projection. Show that if T= tan (theta) then T satisfies the equation:

    75T^2 - 40T + 4 =0

    What is confusing me is how to do this without values as I have an equation to form the quadratic equation:

    x tan(theta) - x^2(g/2u^2)sec^2(theta)
    Here's a partial solution anyway. Assume that \theta is the projection angle. Then if you take the ratio of the y component of the initial velocity and the x component of the initial velocity you get tan(\theta). So write out the x equation of motion and the y equation of motion:
    d = v_0~cos(\theta)~t

    \frac{d}{10} = v_0~sin(\theta)~t - \frac{1}{2}gt^2

    Thus
    tan(\theta) = \frac{v_0~sin(\theta)}{v_0~cos(\theta)} = \frac{\frac{d}{10t} + \frac{1}{2}gt}{\frac{d}{t}}

    tan(\theta) = \frac{1}{10} + \frac{g}{2d}t^2

    From the cosine equation
    t = \frac{d}{v_0~cos(\theta)}

    so
    tan(\theta) = \frac{1}{10} + \frac{g}{2d} \left ( \frac{d}{v_0~cos(\theta)} \right )^2

    tan(\theta) = \frac{1}{10} + \frac{gd}{2v_0^2}~sec^2(\theta)

    tan(\theta) = \frac{1}{10} + \frac{gd}{2v_0^2}~(1 + tan^2(\theta))

    At the moment I can't think of a way to eliminate the v_0 from this.

    -Dan
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    i think it maybe easier than it sounds as the question is only 6 marks and this includes find the values of the quadratic equations.

    This was the previous part of the question:

    An object of mass M is projected from the floor of a horizontal tunnel of height 2d/15 with an initial speed of u at an angle (theta) above the horizontal.

    the object just touches the tunnel roof show that:

    u^2 = 4dg/15sin^2(theta)


    Now ive done that. So can i use this equation to form the quadratic equation???
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by thermalwarrior View Post
    i think it maybe easier than it sounds as the question is only 6 marks and this includes find the values of the quadratic equations.

    This was the previous part of the question:

    An object of mass M is projected from the floor of a horizontal tunnel of height 2d/15 with an initial speed of u at an angle (theta) above the horizontal.

    the object just touches the tunnel roof show that:

    u^2 = 4dg/15sin^2(theta)


    Now ive done that. So can i use this equation to form the quadratic equation???
    Ah, good. That gives us what we need. Your u is my v_0, so continuing:
    tan(\theta) = \frac{1}{10} + \frac{gd}{2v_0^2}sec^2(\theta)

    with
    v_0^2 = \frac{4dg}{15~sin^2(\theta)}

    gives
    tan(\theta) = \frac{1}{10} + \frac{gd}{2 \left ( \frac{4dg}{15~sin^2(\theta)} \right ) }sec^2(\theta)

    tan(\theta) = \frac{1}{10} + \frac{15}{8}sin^2(\theta)~sec^2(\theta)

    tan(\theta) = \frac{1}{10} + \frac{15}{8}tan^2(\theta)

    40~tan(\theta) = 4 + 75~tan^2(\theta)

    or finally:
    75~tan^2(\theta) - 40~tan(\theta) + 4 = 0

    -Dan
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  5. #5
    Senior Member nikhil's Avatar
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    Smile Very simple

    Hi, thermalwarrior.
    This is a very easy question.
    Horizontal range is given by the formula=>[u^2sin(2@)]/g
    and vertical height by the formula =>(usin(@))^2/2g.
    [@=theta and u=initial velocity]
    According to the question
    d=[u^2sin(2@)]/g and
    d/10=(usin(@))^2/2g
    dividing both we get
    tan@=2/5
    [sin2@=2sin@cos@]
    putting value of tan@ in the given expression we get
    75(2/5)^2-40(2/5)+4=0 hence proved.
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