Conservation of energy:

$\overline{)\begin{array}{rcl}\mathbf{P}{\mathbf{E}}_{\mathbf{i}}\mathbf{+}\mathbf{K}{\mathbf{E}}_{\mathbf{i}}& {\mathbf{=}}& \mathbf{P}{\mathbf{E}}_{\mathbf{f}}\mathbf{+}\mathbf{K}{\mathbf{E}}_{\mathbf{f}}\\ \mathbf{m}\mathbf{g}{\mathbf{h}}_{\mathbf{i}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{{\mathbf{v}}_{\mathbf{i}}}^{\mathbf{2}}& {\mathbf{=}}& \mathbf{m}\mathbf{g}{\mathbf{h}}_{\mathbf{f}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{{\mathbf{v}}_{\mathbf{f}}}^{\mathbf{2}}\end{array}}$

v_{i} = v m/s, v_{f} = 0 m/s, h_{i} = 0m, h_{f} = h_{max}

An object is thrown directly upward with some initial speed *v*.

Using conservation of energy, find the maximum height *h*_{max} to which the object will rise.

Express your answer in terms of *v* and the magnitude of the acceleration of gravity *g* .

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