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### Transposition of formulae, is very important/helpfull programming skill, Im stuck

So im improving maths but I cant work out how to get to the correct answer with this algebraic chunk;

with `E=mv2 + mgh` find what v is.

i actually have the answer `2E/v2 + 2gh ` = v

I actually tried to square root it which is wrong.

anyone know the correct way to do this.

plus for those that haven't taken programming to this kind of maths don't be intimidated like I was, Its easier to access this level of maths than you think and its fun especially now that I can recreate pendulums or speed of light equations :)

EDIT: My house mates a good physicist and mathematician (but he is out and i need to move on sooner) help me compile a list of good questions and I might provide a thread with answers and explanations, bonus points for finding a subtle question that will do his head in
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What you do to one side of the equation, you do to the other.

 ``1234567891011121314151617`` ``````E = mv2 + mgh | v E - mgh = mv2 | v (E-mgh) / m = v2 | v (E/m) - gh = v2 | v ((E/m) - gh)1/2 = v``````
thats not the same answer as 2E/v2 + 2gh

i think im trying not to square root it.
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Putting something to the half power is square rooting it.

`X1/2 = sqrt(X)`

On a side note, the formula for the kinetic energy of an object is: `KE = (1/2)mv2`
`E=mv2 + mgh`
and
`2E/v2 + 2gh = v`

Are not the same equation (I forget the term for this .. the two are not equivilent) because 'm' was dropped in the 2nd equation, and there is no way to safely drop 'm' from the first equation unless you 'plug in' a number for it.

wait... E = 1/2mv2+mgh...sry
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Is that the "solution"? Because that's solved for E, not for v.

As I mention the posted solution (in your original post) cannot possibly be the solution for that equation because it lacks 'm'. m cannot just be dropped from the equation.

EDIT:

Also, is that `0.5 * mv2` or `1/(2mv2)`? The way it's written is a little ambiguous.
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the question is solve E=(one half of m)timesv2 + mgh for v

it say that the answer is` v = 2E/v2 + 2gh`

so that is v= 2E with a line under and written under that line is v2 + 2gh

it is the second page third question here at: http://www.mathtutor.ac.uk/algebra/transpositionofformulae/exercise

EDIT: I am stuck.
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 i actually have the answer 2E/v2 + 2gh = v

 wait... E = 1/2mv2+mgh...sry

If you mean to say the answer is 2E/(v2 + 2gh) = m with that original equation, I could buy that. But, that is obviously solved for m and not v.

It helps if you're precise about what you're asking.
okay...rearange this formulae so that it makes the quantity shown the subject

solve E = 1/2mv2+mgh for v.

so v = what?

heres another example...

what is h when you have the formulae A = 2pi r2 + 2pi rh

answer h = A-2pir2/2 pi r

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 it is the second page third question here at: http://www.mathtutor.ac.uk/algebra/transpositionofformulae/exercise

Since that question is an exact duplicate of the one before it, but the correct answer for the one before it isn't accepted as an answer for that one, I'm inclined to believe the question is not what the answer expects it to be. It accepts the answer when the equation is solved for m, so I suspect the v is a typo.
really??

OMG they are identicle!!! no wonder I was stuck for ages !

what could be worse than a miss printed test!
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