1. **Problem statement:** A flowerpot falls past a window 1.9 m tall in 0.20 s. We need to find the height above the top of the window from which the flowerpot was dropped. 2. **Known values:** - Time to fall past window, $t = 0.20$ s - Height of window, $h = 1.9$ m - Acceleration due to gravity, $g = 9.8$ m/s² downward 3. **Concepts and formulas:** - The flowerpot is in free fall, so its motion is uniformly accelerated. - Use the kinematic equation for displacement with initial velocity $v_0$ and acceleration $a$: $$s = v_0 t + \frac{1}{2} a t^2$$ - The flowerpot passes the window in 0.20 s, so the displacement during this time is 1.9 m. 4. **Step 1: Find velocity of flowerpot at the top of the window ($v$):** - Let $v$ be the velocity of the flowerpot as it reaches the top of the window. - During the 0.20 s it falls past the window, it travels 1.9 m with acceleration $g$. - Using displacement formula: $$1.9 = v \times 0.20 + \frac{1}{2} \times 9.8 \times (0.20)^2$$ $$1.9 = 0.20 v + 0.5 \times 9.8 \times 0.04$$ $$1.9 = 0.20 v + 0.196$$ - Solve for $v$: $$0.20 v = 1.9 - 0.196 = 1.704$$ $$v = \frac{1.704}{0.20} = 8.52 \text{ m/s}$$ 5. **Step 2: Find height above the window from which the flowerpot was dropped ($H$):** - The flowerpot started from rest and fell freely under gravity to reach velocity $v$ at the top of the window. - Use the kinematic equation: $$v^2 = v_0^2 + 2 g H$$ Since $v_0 = 0$ (dropped from rest), $$H = \frac{v^2}{2 g} = \frac{(8.52)^2}{2 \times 9.8}$$ $$H = \frac{72.59}{19.6} = 3.71 \text{ m}$$ 6. **Conclusion:** The flowerpot was dropped from a height of approximately **3.71 meters** above the top of the window.