1. **State the problem:** A freely falling object requires 1.50 s to travel the last 30.0 m before it hits the ground. We need to find the height from which it fell. 2. **Define variables and known values:** - Time to fall last 30.0 m: $t_{last} = 1.50$ s - Distance fallen in last interval: $d = 30.0$ m - Acceleration due to gravity: $g = 9.80$ m/s$^2$ 3. **Analyze the motion:** Let the total height fallen be $H$ and the time to fall from rest to the ground be $T$. 4. **Use kinematic equations:** The total distance fallen in time $T$ is $$H = \frac{1}{2} g T^2$$ The distance fallen in time $T - t_{last}$ is $$H - d = \frac{1}{2} g (T - t_{last})^2$$ 5. **Set up equation for last 30 m:** Subtracting, $$d = H - (H - d) = \frac{1}{2} g T^2 - \frac{1}{2} g (T - t_{last})^2$$ Simplify: $$d = \frac{1}{2} g \left(T^2 - (T - t_{last})^2\right)$$ Expand the square: $$T^2 - (T - t_{last})^2 = T^2 - (T^2 - 2T t_{last} + t_{last}^2) = 2T t_{last} - t_{last}^2$$ So, $$d = \frac{1}{2} g (2T t_{last} - t_{last}^2) = g T t_{last} - \frac{1}{2} g t_{last}^2$$ 6. **Solve for $T$:** $$d = g T t_{last} - \frac{1}{2} g t_{last}^2$$ $$g T t_{last} = d + \frac{1}{2} g t_{last}^2$$ $$T = \frac{d + \frac{1}{2} g t_{last}^2}{g t_{last}}$$ Plug in values: $$T = \frac{30.0 + \frac{1}{2} \times 9.80 \times (1.50)^2}{9.80 \times 1.50} = \frac{30.0 + 11.025}{14.7} = \frac{41.025}{14.7} \approx 2.79 \text{ s}$$ 7. **Calculate total height $H$:** $$H = \frac{1}{2} g T^2 = \frac{1}{2} \times 9.80 \times (2.79)^2 = 4.90 \times 7.78 = 38.1 \text{ m}$$ **Final answer:** The object fell from a height of approximately **38.1 meters**.