Question

A rock is thrown upwards and has a max altitude of 40 m. What was its initial velocity?

Answer 1

Recall the formula,

${v_f}^2-{v_i}^2=2a\Delta x$

where $v_i$ and $v_f$ are the rock's initial and final velocities, respectively; $a$ is its acceleration; and $\Delta x$ is the displacement it undergoes.

At any point during its motion, the rock is subject to gravity, so $a=-g$, where $g=9.80\frac{\rm m}{\mathrm s^2}$. At its maximum height, the rock has zero vertical velocity, and if we take its starting height to be the origin, we have $\Delta x=x_{\rm max}$.

So,

$0^2-{v_i}^2=-2gx_{\rm max}\implies-{v_i}^2=-2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(40\,\mathrm m\implies\boxed{v_i=28\dfrac{\rm m}{\rm s}}$