If a rock climber accidentally drops a 56.0 g piton from a height of 295 m, what would its speed be just before striking the ground? Ignore the effects of air resistance.

The speed of the piton before hitting the ground = 76.04 m/s

Explanation:

From the law of conservation of energy,

The kinetic energy of the piton = potential energy of the piton

Note:Before striking the ground, the potential energy of the piton is converted to kinetic energy, with no energy lost to air resistance.

1/2mv² = mgh ... Equation 1

Where m = mass of the piton, v = speed of the piton, h = height, g = acceleration due to gravity.

making v the subject of the equation above,

v = √ (2gh) ... Equation 2

Given: h = 295 m, g = 9.8 m/s²

Substituting into equation 2

v = √ (2*9.8*295)

v = √5782

v = 76.04 m/s

Thus the speed of the piton before hitting the ground = 76.04 m/s