1. **State the problem:** A spring-operated gun shoots a pellet to a maximum height of 100 m on Earth where gravitational acceleration is $g_{Earth} = 9.8$ m/s². We want to find the maximum height $h_{Moon}$ the pellet could reach on the Moon where $g_{Moon} = 1.6$ m/s². 2. **Relevant formula:** The maximum height reached by a projectile under gravity is given by the energy conservation or kinematic formula: $$h = \frac{v^2}{2g}$$ where $v$ is the initial velocity and $g$ is the gravitational acceleration. 3. **Key insight:** The initial velocity $v$ imparted by the spring is the same on Earth and Moon because it depends on the spring, not gravity. 4. **Calculate initial velocity on Earth:** $$v = \sqrt{2gh_{Earth}} = \sqrt{2 \times 9.8 \times 100} = \sqrt{1960}$$ 5. **Calculate height on Moon:** Using the same $v$: $$h_{Moon} = \frac{v^2}{2g_{Moon}} = \frac{2gh_{Earth}}{2g_{Moon}} = \frac{g_{Earth}}{g_{Moon}} \times h_{Earth}$$ 6. **Substitute values:** $$h_{Moon} = \frac{9.8}{1.6} \times 100 = 6.125 \times 100 = 612.5 \text{ m}$$ 7. **Interpretation:** The pellet would rise approximately 610 m on the Moon. **Final answer:** E) 610 m