How to derive this equation of missile launch?

  • Thread starter CarlosK
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  • #1
CarlosK
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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.
 
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Answers and Replies

  • #2
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They seem to be derived for a specific problem statement. How exactly are they used?
 
  • #3
CarlosK
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The exercise asked to deduce these formulas.
 
  • #4
36,027
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Without any other context or diagram? Then the problem statement is really bad.
 
  • #5
Ray Vickson
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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.

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)).
 
  • #6
CarlosK
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Sorry..
v = √ (2gh/(β+1)) is correct. Thanks
 
  • #7
CarlosK
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I got this v = √ (2gh/(β+1)).
I need this v=∆x√(g/(2H))
 
  • #8
CarlosK
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}{2}\;\;\to\;\;t=\sqrt{\frac{2h}{g}}\;\;\;\therefore&space;\;\;\;\frac{1}{t}=\sqrt{\frac{g}{2h}}.gif




atex?v_x=\frac{\Delta&space;x}{t}\;\;\Rightarrow\;\;&space;v_x=\Delta&space;x\sqrt{\frac{g}{2h}}.gif

by: Euclides (http://pir2.forumeiros.com/)
 
  • #9
36,027
12,927
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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