Finding max. height of object thrown vertically upard

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To find the maximum height of an object thrown vertically upward with an initial speed of 47 m/s at two-thirds of its maximum height, the relevant equations are V^2 - Vo^2 = 2gh and V^2 - Vo^2 = 2g(1/3H). The final velocity at the peak height is 0 m/s, and the velocity at two-thirds of the height is 47 m/s. By applying these equations, one can derive the maximum height by manipulating the equations and considering gravitational acceleration as a vector. The solution involves subtracting the equations to isolate variables and solve for the maximum height.
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Homework Statement


An object is thrown vertically upward such that it has a speed of 47 m/s when it reaches two thirds of its maximum height above launch point. Find maximum height h.



Homework Equations



V^2-Vo^2 = 2gh
or
V^2-Vo^2 = 2g(1/3H)?

The Attempt at a Solution


Final velocity= 0 m/s
Velocity at 2/3h= 47 m/s
Acceleration of gravity=9.8 m/s^2
 
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dont confuse yourself .. You can solve the following equations to get the max. Height :

(47)^2 - vo^2 = 2g(2/3) ymax
0 - vo^2 = 2g ymax

You can as a start subtract the second equation from the first .. And keep in mind that you are using g as a vector not a scalar value in that equation ..
 

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