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How to derive this equation of missile launch?

  1. Jun 14, 2016 #1
    Hi All,
    Question: How to derive these equations of missile launch?

    v=∆x√(g/2H)

    v=√(2gh/β+1)

    I need to use these formulas:
    ∆y=voyt - (gt²)/2
    ∆x=vxt
    K=(mv²)/2 + (Iω²)/2
    ∆UG=mg∆y
    I=βmr²
    β=2/5

    Thanks, Carlos.
     
    Last edited: Jun 14, 2016
  2. jcsd
  3. Jun 14, 2016 #2

    mfb

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    They seem to be derived for a specific problem statement. How exactly are they used?
     
  4. Jun 14, 2016 #3
    The exercise asked to deduce these formulas.
     
  5. Jun 14, 2016 #4

    mfb

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    Without any other context or diagram? Then the problem statement is really bad.
     
  6. Jun 14, 2016 #5

    Ray Vickson

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    When parsed according to standard rules for reading mathematical expressions, the second formula above means
    [tex] v = \sqrt{ \frac{2gh}{\beta} + 1}[/tex]
    Is that really what you wanted, or did you mean
    [tex] v = \sqrt{ \frac{2gh}{\beta + 1}} ?[/tex]
    If the latter, you need to use parentheses, like this: v = √ (2gh/(β+1)).
     
  7. Jun 14, 2016 #6
    Sorry..
    v = √ (2gh/(β+1)) is correct. Thanks
     
  8. Jun 14, 2016 #7
    I got this v = √ (2gh/(β+1)).
    I need this v=∆x√(g/(2H))
     
  9. Jun 14, 2016 #8
  10. Jun 14, 2016 #9

    mfb

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    Combining those equations only makes sense if ##\Delta x=h## (they have to apply to the same time t), and if vx is the average velocity during ascent (or descent).
     
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