1. **Problem statement:** A particle is projected horizontally at 20 m/sec from a height of 7.84 m. We need to find the time taken to reach the surface and the horizontal distance travelled in that time. 2. **Relevant formulas:** - Time to fall from height $h$ under gravity $g$ is given by $$t = \sqrt{\frac{2h}{g}}$$ - Horizontal distance travelled is $$d = v_x \times t$$ where $v_x$ is horizontal velocity. 3. **Given values:** - Initial horizontal velocity, $v_x = 20$ m/sec - Height, $h = 7.84$ m - Acceleration due to gravity, $g = 9.8$ m/sec$^2$ 4. **Calculate time to reach the surface:** $$t = \sqrt{\frac{2 \times 7.84}{9.8}} = \sqrt{\frac{15.68}{9.8}} = \sqrt{1.6} = 1.2649 \text{ seconds (approx)}$$ 5. **Calculate horizontal distance travelled:** $$d = 20 \times 1.2649 = 25.298 \text{ meters (approx)}$$ **Final answers:** - Time taken to reach the surface: $1.26$ seconds (rounded to two decimals) - Horizontal distance travelled: $25.3$ meters (rounded to one decimal)