202109160908 Exercises 5.1.A (Q18)

Use the free fall motion equation for position to show that the maximum height reached by an object launched straight up from the ground with an initial velocity $v_0$ is $\frac{v_0^2}{2g}$ .

Proof.

$\begin{aligned} v^2 & = u^2+2as \\ \dots \textrm{ upward} & \textrm{ taken to be positive }\dots \\ 0 & = v_0^2 + 2(-g)s \\ s & = \frac{v_0^2}{2g} \end{aligned}$

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