How to derive this equation of missile launch?

In summary, the conversation discusses deriving equations for missile launch, specifically v=∆x√(g/2H) and v=√(2gh/β+1). The formulas ∆y=voyt - (gt²)/2, ∆x=vxt, K=(mv²)/2 + (Iω²)/2, ∆UG=mg∆y, and I=βmr² are mentioned as well as the value of β being 2/5. There is some discussion about the correct form of the second formula and the need for parentheses.
  • #1
CarlosK
5
0
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:
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  • #2
They seem to be derived for a specific problem statement. How exactly are they used?
 
  • #3
The exercise asked to deduce these formulas.
 
  • #4
Without any other context or diagram? Then the problem statement is really bad.
 
  • #5
CarlosK said:
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
Sorry..
v = √ (2gh/(β+1)) is correct. Thanks
 
  • #7
I got this v = √ (2gh/(β+1)).
I need this v=∆x√(g/(2H))
 
  • #8
}{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
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).
 

Related to How to derive this equation of missile launch?

1. What is the basic equation for calculating missile launch?

The basic equation for calculating missile launch is the Newton's second law of motion, which states that force equals mass times acceleration (F=ma). This equation can be used to determine the initial velocity and acceleration of the missile during launch.

2. How do you determine the initial velocity of a missile during launch?

The initial velocity of a missile can be determined by taking into account the velocity of the missile at the time of launch, the angle of launch, and the acceleration due to gravity. This can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

3. What factors affect the trajectory of a missile during launch?

The trajectory of a missile during launch is affected by various factors such as the initial velocity, angle of launch, air resistance, and the force of gravity. Other external factors such as wind speed and direction can also have an impact on the trajectory of the missile.

4. How does air resistance affect the equation for missile launch?

Air resistance, also known as drag, can have a significant impact on the equation for missile launch. It is a force that opposes the motion of the missile, making it harder for the missile to maintain its velocity. This force can be taken into account by using the drag coefficient in the equation for calculating missile launch.

5. Can the equation for missile launch be used for all types of missiles?

The basic equation for missile launch can be used for most types of missiles, but it may need to be modified depending on the specific design and characteristics of the missile. For example, missiles with wings or fins will experience different amounts of air resistance and will require additional equations to accurately calculate their launch trajectory.

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