Before it hits the sand bed, the meteorite is accelerating uniformly with [tex]g=9.80\dfrac{\rm m}{\mathrm s^2}[/tex], so that its speed [tex]v[/tex] satisfies
[tex]v^2-{v_0}^2=-2g\Delta y[/tex]
where [tex]v_0=90.0\dfrac{\rm m}{\rm s}[/tex] is its initial speed and [tex]\Delta y=(0-850)\,\mathrm{km}=-850\,\mathrm{km}[/tex] is its change in altitude. Notice that we're taking the meteorite's starting position in the atmosphere to be the origin, and the downward direction to be negative. Now,
[tex]v^2-\left(-90.0\dfrac{\rm m}{\rm s}\right)^2=-2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(-850,000\,\mathrm m)\implies\boxed{v=4080\dfrac{\rm m}{\rm s}}[/tex]
