1. **Problem statement:** We want to find the vertical position of the tip of the hour hand of length 12 cm as it rotates over 72 hours, starting at the 9 o'clock position. 2. **Understanding the problem:** The hour hand rotates 360 degrees every 12 hours. The angle $\theta$ in degrees after $t$ hours is given by: $$\theta = 30t + \theta_0$$ where $30$ degrees per hour is the angular speed (360 degrees / 12 hours), and $\theta_0$ is the initial angle. 3. **Initial angle:** At 9 o'clock, the hour hand points to 9, which corresponds to $270^\circ$ from the 12 o'clock position (0 degrees). So, $\theta_0 = 270^\circ$. 4. **Vertical position formula:** The vertical position $y$ of the tip is given by: $$y = r \sin(\theta)$$ where $r=12$ cm is the length of the hour hand, and $\theta$ must be in radians for the sine function. 5. **Convert degrees to radians:** $$\theta_{rad} = \frac{\pi}{180} \theta = \frac{\pi}{180} (30t + 270)$$ 6. **Final formula for vertical position:** $$y(t) = 12 \sin\left(\frac{\pi}{180} (30t + 270)\right)$$ 7. **Time interval:** We consider $t$ from 0 to 72 hours. 8. **Summary:** The vertical position of the tip of the hour hand over 72 hours is: $$y(t) = 12 \sin\left(\frac{\pi}{180} (30t + 270)\right), \quad 0 \leq t \leq 72$$ This function oscillates with period 12 hours, repeating 6 full cycles over 72 hours.