1. **State the problem:** A child on a swing is pulled back to a height of 0.50 m above the equilibrium position. We need to find the speed of the child as they pass through the equilibrium position. 2. **Relevant formula:** We use conservation of mechanical energy. The potential energy at the highest point converts to kinetic energy at the lowest point. Potential energy at height $h$ is given by: $$PE = mgh$$ Kinetic energy at the lowest point is: $$KE = \frac{1}{2}mv^2$$ 3. **Apply conservation of energy:** $$mgh = \frac{1}{2}mv^2$$ Mass $m$ cancels out: $$gh = \frac{1}{2}v^2$$ 4. **Solve for velocity $v$:** Multiply both sides by 2: $$2gh = v^2$$ Take the square root: $$v = \sqrt{2gh}$$ 5. **Substitute values:** Given $g = 9.8$ m/s$^2$, $h = 0.50$ m: $$v = \sqrt{2 \times 9.8 \times 0.50} = \sqrt{9.8}$$ 6. **Calculate:** $$v \approx 3.13 \text{ m/s}$$ **Final answer:** The child is moving at approximately 3.13 m/s when passing through the equilibrium position.