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Time interval btwn 2 particles colliding (each with diff v and elevation)

  1. Oct 18, 2011 #1
    1. The problem statement, all variables and given/known data

    Two particles are projected from the same point with velocities v1 and v2 at elevation α1 and α2, respectively, (α1>α2). Show that if they collide in mid-air, interval between firings must be:

    2v1v2sin(α1 - α2)/g(v1cosα1 + v2cosα2)

    2. Relevant equations

    y = xtanα - (1/2)(gx^2)(sec^2(α))/v^2

    t = (vsinα)/g

    3. The attempt at a solution

    since both particles are colliding, they will have the same x,y value, therefore:

    xtanα1 - (1/2)(gx^2)(sec^2(α1))/v1^2 = xtanα2 - (1/2)(gx^2)(sec^2(α2))/v2^2

    xtanα1 - xtanα2 = (1/2)(gx^2)(sec^2(α2))/v2^2 - (1/2)(gx^2)(sec^2(α1))/v1^2

    x(tanα1 - tanα2) = (1/2)(gx^2)[sec^2(α2)/v2^2 -sec^2(α1)/v1^2]

    (tanα1 - tanα2) = (1/2)(gx)[sec^2(α2)/v2^2 -sec^2(α1)/v1^2]

    and with the other equation:

    t1 = (v1sinα1)/g and t2 = (v2sinα2)/g

    t1 - t1 = (v1sinα1)/g - (v2sinα2)/g

    t1 - t1 = (1/g)[v1sinα1 - v2sinα2]

    Now I'm stuck because I'm not sure how to plug one equation into the other... and by the same step eliminate the x from the equation.

    Any clues would be appreciated! Thanks
  2. jcsd
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