UAM expressions:

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

${\mathbf{}}{\mathit{v}}_{{\mathit{f}}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{v}}_{{\mathbf{0}}}\mathbf{}\mathbf{-}\mathit{g}\mathit{t}$

**V _{0} = 20 m/s**

At t = 0 s, v_{0} = (20) - (10)(0) = **2****0 m/s**

At t = 2 s, v_{0} = (20) - (10)(2) = **0 m/s**

At t = 4 s, v_{0} = (20) - (10)(4) = **-2****0 m/s**

At t = 6 s, v_{0} = (20) - (10)(6) = **-40 m/s**

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 velocity, *v** _{y}*(

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Motion Diagrams concept. If you need more Motion Diagrams practice, you can also practice Motion Diagrams practice problems.