1. **Problem statement:** A small rock is thrown vertically upward with an initial speed of 18.0 m/s from the edge of a building 3000 cm tall. We want to find the speed of the rock just before it hits the street in meters per minute (m/min). 2. **Convert building height to meters:** $$3000\ \text{cm} = \frac{3000}{100} = 30\ \text{m}$$ 3. **Known values:** - Initial velocity, $v_0 = 18.0\ \text{m/s}$ (upward) - Height of building, $h = 30\ \text{m}$ - Acceleration due to gravity, $g = 9.8\ \text{m/s}^2$ (downward) 4. **Use energy conservation or kinematic equations:** The rock goes up, stops momentarily, then falls down past the starting point and hits the ground 30 m below. 5. **Calculate maximum height above the roof:** $$v^2 = v_0^2 - 2gh_{max} \implies 0 = (18.0)^2 - 2 \times 9.8 \times h_{max}$$ $$h_{max} = \frac{(18.0)^2}{2 \times 9.8} = \frac{324}{19.6} = 16.53\ \text{m}$$ 6. **Total height from ground at max point:** $$H = h + h_{max} = 30 + 16.53 = 46.53\ \text{m}$$ 7. **Calculate speed just before hitting the ground using:** $$v = \sqrt{2gH} = \sqrt{2 \times 9.8 \times 46.53} = \sqrt{911.6} = 30.19\ \text{m/s}$$ 8. **Convert speed to meters per minute:** $$30.19\ \text{m/s} \times 60 = 1811.4\ \text{m/min}$$ 9. **Closest answer choice:** 1812 m/min (option d) **Final answer:** 1812 m/min