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Conceptual problem with projectile motion

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img23.imageshack.us/img23/4554/physicshomework.png [Broken]


    2. Relevant equations



    3. The attempt at a solution

    I have the right answer because it is obvious, but I wanted to practice working this out regardless, and I have something slightly different for hmax.

    First, I found the y-component of velocity to be:

    [itex]V_{o}sin\theta_{o} - 9.8t[/itex]

    It follows that at the highest point, that this is equal to 0. So I set it equal to 0 and solved for t.

    [itex]t = \frac{V_{o}sin\theta_{o}}{9.8}[/itex]

    Integrating my y velocity gives y position:

    [itex]y(t) = (V_{o}sin\theta_{o})t - 4.9t^{2}[/itex]

    And solving for our time gives me:

    [itex]h_{max} = V_{o}sin\theta_{o}(\frac{V_{o}sin\theta_{o}}{9.8}) - 4.9(\frac{V_{o}sin\theta_{o}}{9.8})^{2}[/itex]

    Where did I go wrong?

    Thanks!

    EDIT nevermind! They are the same, just had to do some algebra.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 7, 2011 #2

    Dick

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    The first thing you did wrong is writing 9.8 instead of g. If you go back and fix that and do a little algebra you'll see you already have the same answer as in the image.
     
  4. Sep 7, 2011 #3
    I'm not sure why you think that's wrong, I didn't forget halfway through the problem that they are the same..
     
  5. Sep 7, 2011 #4

    Dick

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    It's wrong because now you have to change 9.8 back to g and 4.9 back to g/2. Why didn't you just leave it g to begin with? That's what I mean by 'wrong'. It's sort of a tactical error. You're making life complicated by leaving numbers floating around instead of 'g'.
     
  6. Sep 7, 2011 #5
    Not sure, I just wrote it that way without thinking. What I meant was that I didn't get to the end and say "they have g where I have 9.8, what is going on."
     
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