How to derive this equation of missile launch?

[CarlosK]

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.

 

[mfb]

They seem to be derived for a specific problem statement. How exactly are they used?

 

[CarlosK]

The exercise asked to deduce these formulas.

 

[mfb]

Without any other context or diagram? Then the problem statement is really bad.

 

[Ray Vickson]

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
$$v = \sqrt{ \frac{2gh}{\beta} + 1}$$
Is that really what you wanted, or did you mean
$$v = \sqrt{ \frac{2gh}{\beta + 1}} ?$$
If the latter, you need to use parentheses, like this: v = √ (2gh/(β+1)).

 

[CarlosK]

Sorry..
v = √ (2gh/(β+1)) is correct. Thanks

 

[CarlosK]

I got this v = √ (2gh/(β+1)).
I need this v=∆x√(g/(2H))

 

[CarlosK]

by: Euclides (http://pir2.forumeiros.com/)

 

[mfb]

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