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Projectile motion particle question

  1. Apr 7, 2005 #1
    A particle is projected from the origin 0 with initial velocity v at an angle of elevation [tex] \theta [/tex] and it moves freely under gravity. Find [tex] tan \theta [/tex] if the greatest height reached by the particle is equals to its range on the horizontal plane.

    Using the equations of motion, i have found

    [tex] \frac {v^2_y}{ 2g} = v_x t =v_y - 1/2 gt^2 [/tex]

    but i am unsure of how to manipulate it from here to find [tex] v_y / v_x [/tex].
  2. jcsd
  3. Apr 7, 2005 #2
    Use these

    [tex]v_x = vcos( \theta)[/tex]
    [tex]v_y = vsin( \theta) -gt[/tex]

    [tex]x = vcos (\theta)t[/tex]
    [tex]y = vsin( \theta)t - g \frac{t^2}{2} [/tex]

    Then, the maximal heigth can be determined by realizing that v_y must be zero there. Calculate the time at which this occurs and put it into the formula for y. same goes for x (set y = 0 and calculate the time and put it into x) and then set these two equal to each other

    Last edited: Apr 7, 2005
  4. Apr 9, 2005 #3
    hey thanks for the help marlon, I've gotten it....
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