A metal ball dropped from rest falls 1.0000 m in 0.45173 s with air resistance negligible. What is the local acceleration due to gravity?

1. Identify the knowns: the initial position is 0, the final position is -1.0000 m, the initial velocity is 0 and the time is 0.45173 s. 2. Use position equals initial position plus initial velocity times time plus half the acceleration times time squared. 3. With zero initial velocity this reduces to displacement equals half the acceleration times time squared. 4. Rearranged, the acceleration equals twice the displacement divided by the time squared. 5. Substitute the values: 2 x (-1.0000 m) / (0.45173 s)^2. 6. Since (0.45173 s)^2 = 0.204060 s^2, the acceleration is -9.8010 m/s^2. 7. The magnitude of the acceleration due to gravity is therefore 9.8010 m/s^2. 8. Dropping the factor of two gives 4.9005 m/s^2, which is the common slip. _Source: OpenStax College Physics (CC BY 4.0), Ch 2 "Kinematics", section 2.7 Falling Objects_