If a 10kg rock falls of a 10m cliff, what is its speed right before it hits the ground? show work.

The rock's potential energy just as it rolls off the edge of the cliff is

   (mass) (gravity) (height) = (10kg) (9.8 m/s²) (10 m) = 980 kg-m²/s²

That's exactly the kinetic energy it will have when it hits the ground.

   Kinetic Energy = (1/2) (mass) (speed)²

   980 kg-m²/s² = (1/2) (10kg) (speed)²

Divide each side by  5kg :

   Speed² = (980 kg-m²/s²) / (5 kg) = (196 m/s)²

   Speed = square root of (196 m²/s²)  =  14 m/s

By the way . . . the mass doesn't matter.  Without air resistance
(which is always ignored in problems like this), It doesn't matter
if it's a rock, a feather, a bowling ball, or a bus.  If it falls from
10 m, then it lands at  14 m/s .  That's how gravity works.

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