UAM expressions:

$\overline{)\mathbf{}{{\mathit{v}}}_{{\mathit{f}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{-}}{\mathit{g}}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{y}}{\mathbf{=}}{\mathbf{}}\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{y}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathit{t}}{\mathbf{-}}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{g}}{{\mathit{t}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{{{\mathit{v}}}_{{\mathit{f}}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{}}{{{\mathit{v}}}_{{\mathbf{0}}}}^{{\mathbf{2}}}{\mathbf{}}{\mathbf{-}}{\mathbf{2}}{\mathit{g}}{\mathbf{\u2206}}{\mathit{y}}}$

Solve for V_{0} — consider from t = 0s to the turning point at t = 2 s:

v_{f} = v_{0} - gt

v_{f} = 0 m/s, g = 10 m/s^{2},

v_{0} = v_{f} + gt = 0 + (10)(2) = 20 m/s

**Graphing:**

A stone is thrown upward from the edge of a cliff, reaches its maximum height, and then falls down into the valley below. A motion diagram for this situation is given (Figure 1) , beginning the instant the stone leaves the thrower’s hand.

Construct a graph corresponding to the stone's vertical displacement, *y*(*t*), taking the acceleration due to gravity as exactly 10 m/s^{2}. Ignore air resistance. The unit of time is in seconds and the unit of displacement is in meters. In plotting the points, round-off the coordinate values to the nearest integer.

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