1. **State the problem:** We are given the position function for an object under gravity: $$s(t) = -\frac{1}{2}gt^2 + v_0 t + h_0$$ where $g = 9.8$, $v_0 = 28$, $h_0 = 3$, and we want to find the position at $t=2$ seconds. 2. **Write the formula and substitute known values:** $$s(2) = -\frac{1}{2} \times 9.8 \times (2)^2 + 28 \times 2 + 3$$ 3. **Calculate each term:** $$-\frac{1}{2} \times 9.8 \times 4 = -\frac{1}{2} \times 39.2 = -19.6$$ 4. **Calculate the velocity term:** $$28 \times 2 = 56$$ 5. **Sum all terms:** $$s(2) = -19.6 + 56 + 3$$ 6. **Simplify:** $$s(2) = 39.4$$ **Final answer:** $$\boxed{39.4}$$ This means the object is at 39.4 meters at $t=2$ seconds.