If a rock climber accidentally drops a 69.0 g piton from a height of 305 meters, what would its speed be just before striking the ground? Ignore the effects of air resistance.

Answer 1

PE=KE
mgh=1/2mv^2

Masses cancel out, so you do not need to know it.
solve for v

(2gh)^1/2=v
(2*(9.8m/s^2)(305m))^1/2=v

Answer 2

2992.05