Step 1: Calculate the horizontal range and maximum height for the first stage. The projectile is launched with initial velocity $v_0$ at an angle $\theta$ with gravitational acceleration $g_1 = g$. The initial horizontal velocity is $v_{0x} = v_0 \cos \theta$. The initial vertical velocity is $v_{0y} = v_0 \sin \theta$. The time to reach the peak of the trajectory ($t_p$) is when the vertical velocity becomes zero: $$v_y = v_{0y} - g_1 t_p$$ $$0 = v_0 \sin \theta - g t_p$$ $$t_p = \frac{v_0 \sin \theta}{g}$$ The horizontal distance covered during the first stage ($R_1$) is: $$R_1 = v_{0x} t_p = (v_0 \cos \theta) \left(\frac{v_0 \sin \theta}{g}\right)$$ $$R_1 = \frac{v_0^2 \sin \theta \cos \theta}{g}$$ The maximum height ($H$) reached by the projectile is: $$H = v_{0y} t_p - \frac{1}{2} g_1 t_p^2$$ $$H = (v_0 \sin \theta) \left(\frac{v_0 \sin \theta}{g}\right) - \frac{1}{2} g \left(\frac{v_0 \sin \theta}{g}\right)^2$$ $$H = \frac{v_0^2 \sin^2 \theta}{g} - \frac{v_0^2 \sin^2 \theta}{2g}$$ $$H = \frac{v_0^2 \sin^2 \theta}{2g}$$ Step 2: Calculate the horizontal range for the second stage. At the peak, the projectile's vertical velocity is $0$, and its horizontal velocity remains $v_{0x} = v_0 \cos \theta$. The gravitational field strength changes to $g_2 = 2g$. The projectile falls from height $H$. The time to fall from height $H$ to the ground ($t_f$) is given by: $$H = \frac{1}{2} g_2 t_f^2$$ $$H = \frac{1}{2} (2g) t_f^2$$ $$H = g t_f^2$$ $$t_f = \sqrt{\frac{H}{g}}$$ Substitute the expression for $H$: $$t_f = \sqrt{\frac{1}{g} \left(\frac{v_0^2 \sin^2 \theta}{2g}\right)}$$ $$t_f = \sqrt{\frac{v_0^2 \sin^2 \theta}{2g^2}}$$ $$t_f = \frac{v_0 \sin \theta}{g \sqrt{2}}$$ The horizontal distance covered during the second stage ($R_2$) is: $$R_2 = v_{0x} t_f = (v_0 \cos \theta) \left(\frac{v_0 \sin \theta}{g \sqrt{2}}\right)$$ $$R_2 = \frac{v_0^2 \sin \theta \cos \theta}{g \sqrt{2}}$$ Step 3: Calculate the total range. The total range ($R_{total}$) is the sum of the horizontal distances from both stages: $$R_{total} = R_1 + R_2$$ $$R_{total} = \frac{v_0^2 \sin \theta \cos \theta}{g} + \frac{v_0^2 \sin \theta \cos \theta}{g \sqrt{2}}$$ Factor out common terms: $$R_{total} = \frac{v_0^2 \sin \theta \cos \theta}{g} \left(1 + \frac{1}{\sqrt{2}}\right)$$ Using the trigonometric identity $\sin(2\theta) = 2 \sin \theta \cos \theta$, we can write $\sin \theta \cos \theta = \frac{1}{2} \sin(2\theta)$: $$R_{total} = \frac{v_0^2}{g} \left(\frac{1}{2} \sin(2\theta)\right) \left(1 + \frac{1}{\sqrt{2}}\right)$$ $$R_{total} = \frac{v_0^2 \sin(2\theta)}{2g} \left(1 + \frac{1}{\sqrt{2}}\right)$$ The new total range of the projectile is: $$\boxed{\frac{v_0^2 \sin(2\theta)}{2g} \left(1 + \frac{1}{\sqrt{2}}\right)}$$