A small object is released from rest and falls 1.00 × 10² feet near the surface of the earth. Neglect air resistance.
-Approximately how fast will the object be moving after falling through the 1.00 × 10² feet mentioned?

Question

Quizplus · Accepted Answer

We can use the equation for gravitational potential energy and kinetic energy to solve for the final velocity.

Gravitational potential energy at the top of the fall = mgh = m(32.2 ft/s²)(1.00 × 10¹² ft) = 3.22 × 10¹³m ft²/s²

At the bottom of the fall, all the potential energy has been converted to kinetic energy, so

Kinetic energy at bottom of the fall = 1/2 mv²

Equating the two:

1/2 mv² = 3.22 × 10¹³m ft²/s²

v² = 6.44 × 10¹³ ft²/s²

v = √(6.44 × 10¹³) ft/s

v ≈ 80 ft/s

Therefore, the best choice is C: 80 ft/s.

A Small Object Is Released from Rest and Falls 1