1. The problem is to understand and analyze the function describing vertical motion under gravity on Mars: $$y(t) = -\frac{1}{2}gt^2 + v_0 t + h_0$$ where $g$ is the acceleration due to gravity on Mars, $v_0$ is the initial velocity, and $h_0$ is the initial height. 2. This is a quadratic function in $t$, representing the height $y$ at time $t$. The formula comes from the kinematic equation for uniformly accelerated motion. 3. Important rules: - The coefficient of $t^2$ is negative, so the parabola opens downward. - The vertex of the parabola gives the maximum height. - The roots (if any) give the times when the object hits the ground ($y=0$). 4. To find the vertex time $t_v$, use the formula $$t_v = -\frac{b}{2a} = -\frac{v_0}{2 \times (-\frac{1}{2}g)} = \frac{v_0}{g}$$ 5. The maximum height is then $$y(t_v) = -\frac{1}{2}g \left(\frac{v_0}{g}\right)^2 + v_0 \left(\frac{v_0}{g}\right) + h_0 = -\frac{1}{2}g \frac{v_0^2}{g^2} + \frac{v_0^2}{g} + h_0 = \frac{v_0^2}{2g} + h_0$$ 6. To find when the object hits the ground, solve $$-\frac{1}{2}gt^2 + v_0 t + h_0 = 0$$ 7. Multiply both sides by $-2$ to clear fractions: $$\cancel{-2} \times \left(-\frac{1}{2}gt^2 + v_0 t + h_0\right) = \cancel{-2} \times 0$$ $$gt^2 - 2 v_0 t - 2 h_0 = 0$$ 8. Use the quadratic formula for $t$: $$t = \frac{2 v_0 \pm \sqrt{(2 v_0)^2 + 8 g h_0}}{2 g} = \frac{2 v_0 \pm \sqrt{4 v_0^2 + 8 g h_0}}{2 g}$$ 9. Simplify the square root and denominator: $$t = \frac{2 v_0 \pm 2 \sqrt{v_0^2 + 2 g h_0}}{2 g} = \frac{v_0 \pm \sqrt{v_0^2 + 2 g h_0}}{g}$$ 10. The positive root gives the time when the object returns to ground level. This analysis helps understand the motion of an object under Mars gravity using the given quadratic function.