How Can I Derive x = (sqrt(2h/g))v from h = .5gt^2 and x = vt?

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In summary, to derive the formula x = (sqrt of 2h/g)v from the equations h = .5gt^2 and x = vt, first solve for t in the first equation by taking the positive root, then substitute that expression for t into the formula x = vt. This will result in the desired formula.
  • #1
blue88
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I need to derive this formula:

x = (sqrt of 2h/g)v

from these two equations:

h = .5gt^2

x = vt


Thanks... just can't figure this out!
 
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  • #2
do you mean [tex] \sqrt{\frac{2h}{g}} v [/tex]

Start off with the first equation. Replace what can be replaced in this equation. Rearrange and you're there!
 
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  • #3
yes, this is the formula, it was for a lab so we were using x as a distance
 
  • #4
Rewrite the first equation for t, I took the positive root:

[tex]h = \frac{{gt^2 }}
{2} \Leftrightarrow t^2 = \frac{{2h}}
{g} \Leftrightarrow t = \sqrt {\frac{{2h}}
{g}} [/tex]

All that's left is substituting it in [itex]x = vt[/itex]
 
  • #5
but there is no way to combine the x=vt with it?
 
  • #6
That's what you have to do now. Substitute the expression we found for t in the formula x = vt...
 
  • #7
Yes, there is. Take [itex]x=vt[/itex] and solve for [itex]t[/itex], then substitute.
 
  • #8
Are you kiddin' me, buddy? How about [tex]t = \frac{{x}}
{v} [/tex]
 
  • #9
but i need to derive this formula:

x = (sqrt of 2h/g)v

from these two equations:

h = .5gt^2

x = vt


i still don't understand how to do that
 
  • #10
The complete solution is right here in this thread.
 

FAQ: How Can I Derive x = (sqrt(2h/g))v from h = .5gt^2 and x = vt?

1. How do I derive the x formula from h and v?

To derive the x formula from h and v, you will need to use the equation of motion for an object with constant acceleration: x = h + v2/2a, where x is the displacement, h is the initial height, v is the initial velocity, and a is the acceleration. Rearrange this equation by isolating x, and you will have the formula for x in terms of h and v.

2. What is the significance of deriving the x formula from h and v?

Deriving the x formula from h and v allows you to calculate the displacement of an object in motion given its initial height and velocity. This is useful in many areas of science, such as physics, engineering, and astronomy, where understanding an object's position is important.

3. Can I use this formula for any type of motion?

Yes, the x formula derived from h and v can be used for any type of motion as long as the acceleration is constant. This includes linear motion, projectile motion, and circular motion.

4. Is there a specific method for deriving the x formula from h and v?

Yes, there is a specific method for deriving the x formula from h and v. It involves using the equation of motion and rearranging it to solve for x. This method is commonly taught in introductory physics courses.

5. Are there any other related formulas I should know?

Yes, there are other related formulas that can be derived from h and v, such as the formula for velocity (v = u + at) and the formula for acceleration (a = (v-u)/t). These formulas can also be derived using the same method as the x formula.

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