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3 April, 02:38

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?

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  1. 3 April, 03:03
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    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
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