An astronaut drops a rock from the top of a crater on the moon. When the rock is halfway down to the bottom of the crater, its speed is what fraction of its final impact speed?

Answer: vf1/vf2 = 1 / sqrt (2)

Explanation : on the moon no drag force so we have only the force of gravity. aceleration is g (moon) = 1.62m/s2. the rest is basic kinematics

if the rock travels H to the bottom we can calculate velocity:

vo=0m/s (drops the rock), yo=0

vf*vf = vo*vo+2g (y-yo)

when the rock is halfway y = H/2 so:

vf1*vf1=2*g*H/2 so vf1 = sqrt (gH)

when the rock reach the bottom y=H so:

vf2*vf2=2*g*H so vf2 = sqrt (2gH)

so vf1/vf2 = 1 / sqrt (2)

good luck from colombia