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Initial Vertical Velocity

  1. Sep 16, 2011 #1
    Hi can anyone help me with this formula:


    vertical_displacement_equation_initial_velocity.png

    How do I end up with this equation, thanks in advance!
     
  2. jcsd
  3. Sep 16, 2011 #2

    PeterO

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    It looks like a re-arrangement of one of the 5 standard equations of motion under constant acceleration - which is what you get during vertical motion near the Earths surface [constant acceleration that is]

    EDIT: Though given the symbols used, it looks like there is an error in the transformation - but without the symbols being defined you can't be certain.
     
  4. Sep 16, 2011 #3
    Can I use it to determine the maximum initial velocity?
     
  5. Sep 16, 2011 #4

    PeterO

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    How would you propose to do that?
     
  6. Sep 16, 2011 #5
    I don't know, I'm really stuck on that maximum initial velocity part
     
  7. Sep 16, 2011 #6

    PeterO

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    maximum initial velocity was not part of your original post. What actual question are you trying to solve?
     
  8. Sep 16, 2011 #7
    The question I want to solve is:

    Estimate the maximum "initial velocity" that you can achieve with a regular tennis ball.

    It is classified as a hard question which got me thinking...

    I think that assumptions are a key role here, at the same time I have seen many other equations on how to solve for the initial velocity - but not the maximum.
     
  9. Sep 16, 2011 #8

    PeterO

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    I answered that question in another thread - look around.
     
  10. Sep 16, 2011 #9
    Is it really that simple? Doesn't it have to do with some trigonometric functions?
     
  11. Sep 16, 2011 #10

    PeterO

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    It is that simple - made so by the words "estimate" and "you".
     
  12. Sep 16, 2011 #11
    What do you mean by "you"?

    (Does your answer involve the maximum initial velocity?)
     
  13. Sep 16, 2011 #12

    PeterO

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    The original question was

    Estimate the "maximum velocity" you can throw a standard Tennis Ball.
     
  14. Sep 16, 2011 #13
    Why do you think this question s classified as hard? Might it be that one should make assumptions? Or what do you think?
     
  15. Sep 16, 2011 #14

    PeterO

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    It is hard because you are not just plugging numbers into a formula. You have to realistically estimate what you yourself can do.
    Since you are throwing a tennis ball, the ball is quite irrelevant as your arm if far heavier than the ball.
    It would be interesting to turn up to call in an upper body cast, and offer the answer 0.1 m/s.

    Another classic example is to hang masses of 100g, 200g, 300g, etc on the end of a spring, and note that it extends say 5cm, 10cm then 15 cm.
    The question is then posed: what if we hang 250kg on the spring.
    Some people suggest 125 m, which follows the nature of the spring, but ignores the fact that the spring is probably made of only 1 metre of wire, coiled up, so is unlikely to become 125 m long.
     
  16. Sep 16, 2011 #15
    What is the vertical motion formulae?
     
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