A rock is thrown straight down from a cliff edge at 13.0 m/s. What is its velocity when it is 5.10 m below the release point?

1. Identify the knowns: the initial velocity is -13.0 m/s, the displacement is -5.10 m and the acceleration is -9.80 m/s^2. 2. All three are negative because upward has been taken as the positive direction. 3. Use the relation that links velocity, acceleration and displacement without time. 4. Substitute the values: (-13.0 m/s)^2 + 2 x (-9.80 m/s^2) x (-5.10 m). 5. This gives 169 m^2/s^2 + 99.96 m^2/s^2 = 268.96 m^2/s^2. 6. Take the square root, which yields plus or minus 16.4 m/s. 7. The rock is still heading down, so the negative root is chosen and the velocity is -16.4 m/s. 8. Simply adding the two speeds gives 22.6 m/s, which is not how the relation works. _Source: OpenStax College Physics (CC BY 4.0), Ch 2 "Kinematics", section 2.7 Falling Objects_