# Chapter 2 - Kinematics in One Dimension - Problems - Page 50: 55

5.06 m downward

#### Work Step by Step

Down is the negative direction: $x= vt - \frac{1}{2}at^{2}$ $x= (-10.1)(1.20) - \frac{1}{2}(-9.8)(1.20)^{2}$ $x=-5.064$ $x\approx-5.06$ Hence, the displacement in the final 1.20s is 5.06 m downward

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.