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

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The formula x = (sqrt(2h/g))v can be derived from the equations h = 0.5gt² and x = vt. By rearranging the first equation to solve for t, we find t = sqrt(2h/g). Substituting this expression for t into the second equation x = vt yields the desired formula. This derivation is essential for understanding the relationship between distance, height, and time in physics.

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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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do you mean \sqrt{\frac{2h}{g}} v

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

h = \frac{{gt^2 }}<br /> {2} \Leftrightarrow t^2 = \frac{{2h}}<br /> {g} \Leftrightarrow t = \sqrt {\frac{{2h}}<br /> {g}}

All that's left is substituting it in x = vt
 
but there is no way to combine the x=vt with it?
 
That's what you have to do now. Substitute the expression we found for t in the formula x = vt...
 
Yes, there is. Take x=vt and solve for t, then substitute.
 
Are you kiddin' me, buddy? How about t = \frac{{x}}<br /> {v}
 
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.
 

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