If a rock climber accidentally drops a 52.5-g piton from a height of 325 meters, what would its speed be just before striking the ground? ignore the effects of air resistance.

Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.

So we have 1/2 MV^2 = MGH

V^2 = 2GH

V = âš2GH

V = âš (2 * 9.8 * 325)

V = âš 6370

V = 79.81 m/s