1. **Problem statement:** A 9 kg particle is on a smooth table connected by a light string over a smooth pulley to a 1 kg particle hanging vertically. The system is released from rest, and the 1 kg particle takes 2 seconds to hit the ground. We need to find the horizontal distance the 9 kg particle moves on the table. 2. **Known values:** - Mass of particle on table, $m_1 = 9$ kg - Mass of hanging particle, $m_2 = 1$ kg - Time for $m_2$ to hit ground, $t = 2$ s - Acceleration due to gravity, $g = 9.8$ m/s$^2$ 3. **Step 1: Find the acceleration of the system.** The system accelerates because of the difference in weights. The acceleration $a$ is given by: $$a = \frac{m_2 g}{m_1 + m_2}$$ 4. **Calculate acceleration:** $$a = \frac{1 \times 9.8}{9 + 1} = \frac{9.8}{10} = 0.98 \text{ m/s}^2$$ 5. **Step 2: Find the distance the 1 kg particle falls.** Using the equation for distance under constant acceleration from rest: $$s = \frac{1}{2} a t^2$$ 6. **Calculate falling distance:** $$s = \frac{1}{2} \times 0.98 \times (2)^2 = 0.49 \times 4 = 1.96 \text{ m}$$ 7. **Step 3: Find the horizontal distance the 9 kg particle moves.** Since the string is inextensible, the 9 kg particle moves horizontally the same distance $s = 1.96$ m. **Final answer:** The 9 kg particle moves horizontally **1.96 meters** from the table edge.