Recall the formula,
[tex]{v_f}^2-{v_i}^2=2a\Delta x[/tex]
where [tex]v_i[/tex] and [tex]v_f[/tex] are the rock's initial and final velocities, respectively; [tex]a[/tex] is its acceleration; and [tex]\Delta x[/tex] is the displacement it undergoes.
At any point during its motion, the rock is subject to gravity, so [tex]a=-g[/tex], where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex]. At its maximum height, the rock has zero vertical velocity, and if we take its starting height to be the origin, we have [tex]\Delta x=x_{\rm max}[/tex].
So,
[tex]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}}[/tex]
