Problem: 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/s2. 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.

FREE Expert Solution

UAM expressions:
$$\begin{gathered} \boxed{\mathbf{ }{\mathit{v}}_{\mathit{f}}\mathbf{ }\mathbf{=}\mathbf{ }{\mathit{v}}_{\mathbf{0}}\mathbf{ }\mathbf{-}\mathit{g}\mathit{t} \\ \mathbf{∆}\mathit{y}\mathbf{=}\mathbf{ }\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}\mathit{t} \\ \mathbf{∆}\mathit{y}\mathbf{=}\mathbf{ }{\mathit{v}}_{\mathbf{0}}\mathit{t}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{g}{\mathit{t}}^{\mathbf{2}} \\ \mathbf{ }{{\mathit{v}}_{\mathit{f}}}^{\mathbf{2}}\mathbf{=}\mathbf{ }{{\mathit{v}}_{\mathbf{0}}}^{\mathbf{2}}\mathbf{ }\mathbf{-}\mathbf{2}\mathit{g}\mathbf{∆}\mathit{y}} \end{gathered}$$

Solve for V₀ — consider from t = 0s to the turning point at t = 2 s:

v_f = v₀ - gt

v_f = 0 m/s, g = 10 m/s², 

v₀ = v_f + gt = 0 + (10)(2) = 20 m/s 

Graphing:

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