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Angular Momentum of Shot Question

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

    In the figure below, a 0.360 kg ball is shot directly upward at initial speed 37.5 m/s.

    hrw7_11-41.gif

    What is its angular momentum about P, 2.15 m horizontally from the launch point, when the ball is at the following heights?
    • a) its maximum height
    • b) halfway back to the ground

    2. Relevant equations

    [itex]\stackrel{\rightarrow}{l} = m(\stackrel{\rightarrow}{r} \times \stackrel{\rightarrow}{v})[/itex], which given this problem simplifies to [itex]l = r_{\bot}mv[/itex]
    [itex]v^2 = v^{2}_{0} + 2a(x - x_{0})[/itex]

    3. The attempt at a solution

    For a), v was obviously 0 at max height, so the angular momentum was 0 as well.

    For b):
    First, I needed to find the halfway position in regards to y, to help find the velocity there. Taking:
    * [itex]v_{initial}[/itex] = the time just after being launched from the ground
    * [itex]v_{final}[/itex] = the apex of its climb

    I plugged in the numbers: [itex]0^2 = 37.5^2 + 2(-9.8)(y - 0)[/itex] and got: [itex] y = 71.747[/itex]

    Now using the same equation, except with [itex]v_{final}[/itex] being halfway up and y obviously half of what it was:
    [itex]v^2 = 0^2 + 2(9.8)(35.874 - 0)[/itex]

    I got [itex]v^2 = 703.125[/itex] so v = -26.517 m/s.

    Edit: Plug in the numbers, and [itex](-26.517 m/s)(2.15 m)(0.36 kg) = 20.52 [/itex] N*m*s, which is right.

    Not sure what I did the first time, but thanks, I guess having to do the post format for this forum helps, haha!
     
    Last edited: Nov 1, 2011
  2. jcsd
  3. Nov 1, 2011 #2

    Chi Meson

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    Glad we could help!
     
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