A rock is thrown straight up from a cliff edge with an initial velocity of 13.0 m/s. Taking upward as positive and the release point as zero, what is its position 1.00 s later?

1. Identify the knowns: the initial position is 0, the initial velocity is +13.0 m/s and the acceleration is -9.80 m/s^2. 2. Upward is positive, so the initial velocity and the acceleration carry opposite signs. 3. Use position equals initial position plus initial velocity times time plus half the acceleration times time squared. 4. Substitute the values: 0 + (13.0 m/s)(1.00 s) + 0.5 x (-9.80 m/s^2)(1.00 s)^2. 5. This gives 13.0 m - 4.90 m = 8.10 m. 6. Quoting 13.0 m forgets the gravity term entirely. 7. Quoting 4.90 m keeps only the gravity term and drops the throw. _Source: OpenStax College Physics (CC BY 4.0), Ch 2 "Kinematics", section 2.7 Falling Objects_