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Object thrown vertically 
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#1
Sep2404, 02:07 PM

P: 36

An object is thrown vertically upward such that it has a speed of 58 m/s when it reaches two thirds of it's maximum height above the launch point.
The acceleration of gravity is 9.8 m/s^2 Find the maximum height h. Answer in units of m. I know: V: 58 m/s at y=2/3 H H = ? V = 0 V^2Vo^2 = 2g(1/3H) I know I need to plug in the numbers, but how do I get all of the equation to one side, to find H?? 


#2
Sep2404, 02:24 PM

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P: 41,568

For example, if you had an equation: 7x^2 = 67bZ, and you wanted to isolate Z, just divide both sides by 67b: (7x^2)/67b = (67bZ)/67b = Z; so Z = (7x^2)/67b. Make sense? 


#3
Sep2404, 02:26 PM

HW Helper
P: 2,277

You got
[tex] V^2 = V_{o}^2 + 2g\frac{2}{3}H [/tex] where V is 58 m/s and [tex] 0 = V_{o}^2 + 2gH [/tex] 


#4
Sep2404, 02:28 PM

HW Helper
P: 2,277

Object thrown vertically
Doc Al, how come initial speed = 0 ?



#5
Sep2404, 02:35 PM

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P: 41,568




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