A body is thrown vertically upward with initial speed

A body is thrown vertically upward with initial speed $v_{0}$. Find (a) its height after $t$ seconds, (b) its velocity and acceleration, (c) the time to reach the greatest height and that height.

Step 1: Set up

Measure $s$ upward from the launch point. The only acceleration is gravity, directed downward: $w = \frac{d^{2}s}{dt^{2}} = -g$.

Step 2: Integrate once

With $v(0) = v_{0}$: $v(t) = v_{0}-gt$.

Step 3: Integrate again

With $s(0) = 0$: $s(t) = v_{0}t-\frac{gt^{2}}{2}$.

Step 4: The highest point

The body rises while $v>0$ and falls when $v\lt 0$, so the summit is where $v = 0$: $v_{0}-gt^{*} = 0 \;\Longrightarrow\; t^{*} = \frac{v_{0}}{g}$.

\[ s_{\max} = v_{0}\cdot\frac{v_{0}}{g}-\frac{g}{2}\cdot\frac{v_{0}^{2}}{g^{2}} = \frac{v_{0}^{2}}{g}-\frac{v_{0}^{2}}{2g} = \frac{v_{0}^{2}}{2g}. \]

Step 5: Verification

\[\text{Energy: } \frac{v_{0}^{2}}{2} = g\,s_{\max} \;\Longrightarrow\; s_{\max} = \frac{v_{0}^{2}}{2g} \;\;\checkmark\]

\[s = 0 \text{ again at } t = \frac{2v_{0}}{g} = 2t^{*}, \text{ with } v = -v_{0}: \text{ rise and fall are mirror images} \;\;\checkmark\]

\[v_{0} = 20\ \text{m/s}:\ t^{*} \approx 2.04\ \text{s},\ s_{\max} \approx 20.4\ \text{m} \;\;\checkmark\]