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(\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(\rm m)/(\mathrm s^2)\right)(40\,\mathrm m\implies\boxed{v_i=28(\rm m)/(\rm s)}`