1. **Problem statement:** Repeat part a for an object thrown upwards on Mars where gravity $g = 3.7 \text{ m/s}^2$. 2. **Recall the formula for maximum height:** $$h = \frac{v_0^2}{2g}$$ where $v_0$ is the initial velocity and $g$ is the acceleration due to gravity. 3. **Important rules:** - Gravity on Mars is less than on Earth, so the object will reach a higher maximum height for the same initial velocity. - The formula assumes upward motion against constant gravity until velocity becomes zero. 4. **Intermediate work:** Assuming the initial velocity $v_0$ is the same as in part a (not given here, so let's denote it as $v_0$), substitute $g = 3.7$: $$h = \frac{v_0^2}{2 \times 3.7} = \frac{v_0^2}{7.4}$$ 5. **Explanation:** The maximum height is inversely proportional to gravity. Since Mars has lower gravity, the object reaches a higher height compared to Earth. 6. **Final answer:** $$\boxed{h = \frac{v_0^2}{7.4}}$$ This is the maximum height reached by the object thrown upwards on Mars with initial velocity $v_0$.