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?

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

1/√2

Explanation:

the rock has lost half of its gravitational potential energy, its kinetic energy at the halfway point is half of its kinetic energy at impact. Since K.E is proportional to v^2, if K.E at halfway point is equal to 1/2 x K.E,

Then,

The K.E at impact, then the rock’s speed at the halfway point

its speed at impact will be 1/√2 its impact speed

pe = weightxheight = 2000x1000=2,000,000joules

small = small current large mag large current