Question

A meteorite has a speed of 90.0 m/s when 850 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.25 m. (a) What is its speed just before striking the sand?

Answer 1

Before it hits the sand bed, the meteorite is accelerating uniformly with $g=9.80\dfrac{\rm m}{\mathrm s^2}$, so that its speed $v$ satisfies

$v^2-{v_0}^2=-2g\Delta y$

where $v_0=90.0\dfrac{\rm m}{\rm s}$ is its initial speed and $\Delta y=(0-850)\,\mathrm{km}=-850\,\mathrm{km}$ 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,

$v^2-\left(-90.0\dfrac{\rm m}{\rm s}\right)^2=-2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(-850,000\,\mathrm m)\implies\boxed{v=4080\dfrac{\rm m}{\rm s}}$