Posted by **john** on Friday, September 25, 2009 at 12:35am.

a boy throws a stone straight upward with an initial speed of 15.0m/s. what mazimum height will the stone reach before falling back down.

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**DrBob222**, Friday, September 25, 2009 at 12:49am
Can you use V^2 = Vo^2 + 2a*distance and set V^2 = 0?

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**Quidditch**, Friday, September 25, 2009 at 12:53am
One method to solve this is conservation of energy.

initial kinetic energy = final potential energy

let m = mass of stone

let g = acceleration of gravity

let v = initial velocity

let h = maximum height

(1/2)mv^2 = mgh

So,

v^2 = 2gh

h = (v^2)/g

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**Anonymous**, Monday, October 3, 2011 at 7:21pm
23

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**Anonymous**, Thursday, September 11, 2014 at 7:58pm
22.95

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**blehh**, Tuesday, September 16, 2014 at 1:42am
final velocity squared = initial velocity squared + 2(acceleration)(displacement)

0 = (15*15) + 2(9.8)x

0 = 225 + 19.6x

-225 = 19.6x

x= 11m

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