Chapter 13 - Oscillations About Equilibrium - Problems and Conceptual Exercises - Page 446: 22

$-0.62m,0.61m,-0.59m$

The x component of the ball's position at the given times can be determined as follows: We know that $x=Acos(\omega t)$.......eq(1) We plug in the known values to obtain: $x=(0.62)cos(1.3\times 2.5)=-0.62m$ Substitute $t=5.0s$ in eq(1) $x=(0.62)cos(1.3\times 5.0)=0.61m$ Substitute $t=7.5s$ in eq(1) $x=(0.62)cos(1.3\times 7.5)=-0.59m$