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Dynamics Problem

  1. Feb 6, 2006 #1
    Bit stuck on a projectile style question, at first glance I thought it would be pretty easy but I'm stuck now!

    "A man is shooting another man who has climbed a pole h metres high, L metres away. The bullets are at speed v leaving the gun. Find an equation determining the angle of projection needed (alpha). Show that tan(alpha)=(v^2)gL for the maximum range."

    Think I've got the angle part sorted, I have arccos(L/vt) for it anyway! But I'm really struggling getting the next part of the question, the nearest I've got is sin(2alpha) = (gL)/v^2 but can't really get tan into it! Any help would be much appreciated. Thanks.
  2. jcsd
  3. Feb 6, 2006 #2
    Horizontal velocity, displacement and acceleration respectively:
    [tex]v_x = v_0cos\alpha[/tex]
    [tex]s_x = v_0tcos\alpha[/tex]
    [tex]a_x = 0[/tex]

    Vertical Velocity, displacement, acceleration respectively:
    [tex]v_y = v_osin\alpha - gt [/tex]
    [tex]s_y = v_0tsin\alpha - \frac{1}{2}gt^2 [/tex]
    [tex]a_y = -g [/tex]

    [tex]\therefore y = \frac{v_0xsin\alpha}{v_0cos\alpha} - \frac{1}{2}g[\frac{x}{v_0cos\alpha}]^2 [/tex]

    Cancel down and re-arrange, and you should be able to get the answer. Just is case it is needed, [tex]sin2\alpha = 2sin\alpha.cos\alpha [/tex] and [tex]tan\alpha = \frac{sin\alpha}{cos\alpha} [/tex]
    I trust that you can finish.
    Last edited by a moderator: Feb 6, 2006
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