1. **Problem statement:** We want to find the time $t$ it takes for an object to fall from a height of 0.55 m under gravity. 2. **Formula used:** The vertical displacement under constant acceleration is given by $$y_f = y_i + V_{iy} t + \frac{1}{2} a_y t^2$$ where $y_f$ is the final position, $y_i$ is the initial position, $V_{iy}$ is the initial vertical velocity, $a_y$ is the vertical acceleration, and $t$ is time. 3. **Given values:** - $y_f = 0$ (ground level) - $y_i = 0.55$ m (initial height) - $V_{iy} = 0$ (starting from rest) - $a_y = -9.81$ m/s$^2$ (acceleration due to gravity, downward) 4. **Substitute values:** $$0 = 0.55 + 0 \times t + \frac{1}{2} (-9.81) t^2$$ which simplifies to $$0 = 0.55 - 4.905 t^2$$ 5. **Isolate $t^2$:** $$-0.55 = -4.905 t^2$$ $$\frac{\cancel{-0.55}}{\cancel{-4.905}} = t^2$$ $$t^2 = 0.1121$$ 6. **Solve for $t$:** $$t = \sqrt{0.1121} = 0.335$$ seconds (approximately) 7. **Interpretation:** The time to fall 0.55 m under gravity starting from rest is about 0.335 seconds. **Your calculation is correct except for the final decimal precision; the time squared is about 0.1121, not exactly 0.11, and the time is about 0.335 s.**