1. **Stating the problem:** We are given the position function of an object moving under constant acceleration: $$s(t) = -\frac{1}{2}gt^2 + v_0 t + h_0$$ where $g$ is the acceleration due to gravity, $v_0$ is the initial velocity, and $h_0$ is the initial height. 2. **Formula and explanation:** This formula comes from the equations of motion under constant acceleration: $$s(t) = s_0 + v_0 t + \frac{1}{2} a t^2$$ Here, acceleration $a = -g$ (negative because gravity acts downward). 3. **Intermediate work:** The formula is already simplified. If you want to find the position at a specific time $t$, plug in the values of $g$, $v_0$, $h_0$, and $t$. 4. **Example:** If $g = 9.8$, $v_0 = 28$, $h_0 = 0$, and $t = 2$ seconds, then $$s(2) = -\frac{1}{2} \times 9.8 \times 2^2 + 28 \times 2 + 0 = -\frac{1}{2} \times 9.8 \times 4 + 56 = -19.6 + 56 = 36.4$$ 5. **Explanation:** This means after 2 seconds, the object is 36.4 meters above the initial height. This formula helps us predict the position of an object under gravity at any time $t$.