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Thread: interception

  1. #16
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    Re: interception

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  2. #17
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    Re: interception

    Vdc =(15cosa-15)i+(15sina-15√3)j
    if you look at skeeter diagram this is this represents the line labelled △r
    $V_{dc}$ is the relative velocity vector, $\color{red}{V_r}$ ... the relative motion (position) vector, $\color{red}{\Delta r}$ is parallel to $\color{red}{V_r}$.


    $\cos{\theta} + \sqrt{3}\sin{\theta} = 1$

    $\sqrt{3}\sin{\theta} = 1-\cos{\theta}$

    $3\sin^2{\theta} = 1 - 2\cos{\theta} + \cos^2{\theta}$

    $3(1-\cos^2{\theta}) = 1 - 2\cos{\theta} + \cos^2{\theta}$

    $0 = 2\cos^2{\theta} - \cos{\theta} - 1$

    $0 = (2\cos{\theta} + 1)(\cos{\theta} - 1) = 0$

    $\cos{\theta} = -\dfrac{1}{2} \implies \theta = 120^\circ$
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