A rock, dropped from rest near the surface of an atmosphere-free planet, attains a speed of 20.0 m/s after falling 8.0 meters.
-What is the magnitude of the acceleration due to gravity on the surface of this planet?

Question

Quizplus · Accepted Answer

Using the kinematic equation $ v^2 = u^2 + 2as $, where $ v = 20.0 \, \text{m/s} $, $ u = 0 \, \text{m/s} $, and $ s = 8.0 \, \text{m} $, we solve for $ a $: $ 20^2 = 0 + 2a(8) $, giving $ a = 25 \, \text{m/s}^2 $.

A Rock, Dropped from Rest near the Surface of an Atmosphere-Free