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Can anyone help me out with this problem: A ball is thrown vertically upward with initial velocity v. Find the maximum height H of the ball as a function of v. Then find the initial velocity v required to achieve a height of H.

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we'll just use the eq, \(v^2 = u^2 +2aS\), since u=0, a=-g,(since it is opposite direction of v), rearranging gives \(S=\frac{v^2}{2g}\)
where S=H
ive arrived at this answer too...however when i submitted said that g was not defined in this context

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Other answers:

is there any way to answer this in terms of H and v only?
or, you can just sub g=10 into it
thats what i would think so too. but it still says that answer is incorrect
oh oh. sorry. my mistake. it's initial velocity, not end velocity. since v=0, \(u=\sqrt{2gH}=\sqrt{19.8H}\)
tried that one too not sure what they looking for
dang==then i have no idea what they want
thanks for trying to help though. ...its very much appreciated :)
you're welcome :)

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