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2 cinematic problems.

  1. Feb 9, 2006 #1

    rea

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    Ok, after some work, it seem that Im not able to do this exercises, over even get something that describe them...

    I dont know even how to handle it... I can see what is asked in some way, but dont know what to plug.


    The next one, I'm more near, but unable to continue.

    here is what I have so far:
    Deducing...
    [tex]a_y = -g [/tex]
    [tex]v_y = -gt+V_0 [/tex]
    [tex]y = -\frac{1}{2}gt^2+v_0t [/tex]

    If [tex]V_y [/tex] eq 0 then the particle is at [tex]y_{max} [/tex] and [tex]t_{max} [/tex] is the time for get there...

    [tex]0 = -gt_{max} + v_0 [/tex]
    [tex]t = \frac{v_0}{g} [/tex]

    then
    [tex]y_{max} = -\frac{1}{2}g\left(\frac{v_0}{g}\right)^2+\frac{v_0v_0}{g} [/tex]
    [tex] = \frac{v_0^2}{2g} [/tex]

    [/tex]

    Now follow a new problem, where d = h + [tex]y_{max}[/tex]
    With start conditions...

    [tex]
    a = g [/tex]
    [tex]v_0=0 [/tex]
    [tex]t_0=0 [/tex]
    [tex]p_0 = 0 [/tex]

    And in the other end of the line:
    [tex]a=g [/tex]
    [tex]v_f = ? [/tex]
    [tex]t_f = ? [/tex]
    [tex]p_f = d = h + y_{max} [/tex]


    I know that this is easy, but now what?
     
    Last edited: Feb 9, 2006
  2. jcsd
  3. Feb 9, 2006 #2
    Yeah. Just set y=0 and solve from there. Use y=y0 + V0T + (1/2)aT^2. Solve for T.

    I think you're forgetting some information in the other problem.
     
  4. Feb 9, 2006 #3

    rea

    User Avatar

    I only dont put the answer that need be proved (I write it in short).

    Perhaps my translatio to english in the question is wrong...

    Here is again:

    calculate the shortes (the littlest) angle that the final velocity of a fragment will do with the horizontal. Solution [tex] tan^{-1}\left[\left(2gh\right)^{\frac{1}{2}} / v \right] [/tex].

    Perhaps that.
     
    Last edited: Feb 9, 2006
  5. Feb 10, 2006 #4

    lightgrav

    User Avatar
    Homework Helper

    If a piece leaves the explosion traveling with its velocity straight DOWN,
    what is the angle that its final velocity makes with the ground?

    If a different piece leaves the explosion with initial velocity straight UP,
    what is the angle that its final velocity makes with the ground?

    If some piece leaves the explosion with its initial velocity HORIZONTAL,
    what is the angle that its final velocity makes with the ground?

    Which final velocity angle is smallest?
    Can you find the final velocity angle as a function of initial velocity angle?
    If so, you can show that horizontal initial velocity should give the smallest final angle, if it is not obvious already (you do want the largest horizontal component, right? and the smallest vertical component...) set [tex]\frac{d\theta_{final}}{d\theta_{init}} = 0[/tex]
     
    Last edited: Feb 10, 2006
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