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Momentum homework but a velocity question

  1. Feb 13, 2008 #1
    This is momentum homework but a velocity question. I can't even figure out where to start with this one. The question is "For your answer for question E (which was V=233m/s) 70% of that velocity goes upward and 70% goes horizontally (45 degree angle). How far will the baseball go (ignoring air resistance)?

    I am not given time so not really sure how to do this one. I was going to try the distance equations of D=1/2A*t^2 and D=avg V*T but I don't have enough info to use those.

    Can anyone tell me where to start?
    Thanks
     
    Last edited: Feb 13, 2008
  2. jcsd
  3. Feb 13, 2008 #2

    Doc Al

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    You can figure out the time. You know the initial velocity in the vertical direction, so how long does it take to reach max height? To fall back down to the ground?
     
  4. Feb 13, 2008 #3
    So the initial velocity in the vertical direction is 163m/s. Do I need to factor in gravity? So would I take 163m/s / 10m/s? So the time to reach it's max height would be 16.3 seconds? I don't think that sound right
     
  5. Feb 13, 2008 #4

    Doc Al

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    Of course you must factor in gravity! But you're doing fine, using v=gt. Sounds good to me. Keep going.
     
  6. Feb 13, 2008 #5
    So since the volocity of 163m/s is the same in the horizontal direction would I do the same thing? V=gt?
     
  7. Feb 13, 2008 #6

    Doc Al

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    No. Gravity only affects the vertical motion. What's the total time the baseball is in the air? How fast does it move horizontally? How far does it get?
     
  8. Feb 13, 2008 #7
    total time would be 32.6s? It's going 163m/s horizontally so would it be D=avg v*t?
     
  9. Feb 13, 2008 #8

    Doc Al

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    Sounds good to me.
     
  10. Feb 13, 2008 #9
    So the total distance is 5313.8m? That seems really far
     
  11. Feb 13, 2008 #10

    Doc Al

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    It's moving pretty fast. And we're neglecting air resistance.
     
  12. Feb 13, 2008 #11
    ok thanks for all your help!
     
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