1. **State the problem:** A rock is thrown downward from a 40.0 m tall tower with an initial speed of 12 m/s. We need to find the speed of the rock just before it hits the ground, assuming negligible air resistance and acceleration due to gravity $g = 9.80$ m/s$^2$. 2. **Formula used:** We use the kinematic equation for velocity under constant acceleration: $$v^2 = v_0^2 + 2 g h$$ where: - $v$ is the final velocity, - $v_0$ is the initial velocity (12 m/s downward), - $g$ is acceleration due to gravity (9.80 m/s$^2$), - $h$ is the height fallen (40.0 m). 3. **Explanation:** Since the rock is thrown downward, initial velocity and acceleration are in the same direction, so we add the terms. 4. **Calculate:** $$v^2 = 12^2 + 2 \times 9.80 \times 40.0$$ $$v^2 = 144 + 784 = 928$$ 5. **Find $v$:** $$v = \sqrt{928} \approx 30.46 \text{ m/s}$$ 6. **Interpretation:** The speed of the rock just before hitting the ground is approximately 30.5 m/s downward.