A projectile is launched with velocity at an angle . At the peak of its trajectory, the gravitational field strength instantly changes from to . What is the new total range of the projectile? (Requires breaking the motion into two distinct, connected stages).

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:

$$H = v_{0y} t_p - \frac{1}{2} g_1 t_p^2$$ $$H = (v_0 \sin \theta) \left(\frac{v_0 \sin \theta}{g}\right) - \frac{1}{2} g \left(\frac{v_0 \sin \theta}{g}\right)^2$$ $$H = \frac{v_0^2 \sin^2 \theta}{g} - \frac{v_0^2 \sin^2 \theta}{2g}$$ $$H = \frac{v_0^2 \sin^2 \theta}{2g}$$

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:

$$R_2 = v_{0x} t_f = (v_0 \cos \theta) \left(\frac{v_0 \sin \theta}{g \sqrt{2}}\right)$$ $$R_2 = \frac{v_0^2 \sin \theta \cos \theta}{g \sqrt{2}}$$

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:

$$\boxed{\frac{v_0^2 \sin(2\theta)}{2g} \left(1 + \frac{1}{\sqrt{2}}\right)}$$