Portal Physics (Part 2)

In the second frame, where the ball emerges from Portal $1$ and enters the center of Portal $2$ , the ball has an initial vertical velocity of zero which means its vertical distance is given by $D = \frac{1}{2}gt^2$ , which solves to $t = \sqrt{\frac{2D}{g}}$ , and its vertical velocity when it reaches Portal $2$ is given by $v_y = gt$ , which solves to $v_y = \sqrt{2Dg}$ . In this same amount of time $t$ , the ball will also travel a horizontal distance $D$ at a constant horizontal velocity $v_x$ at $v_x = \frac{D}{t}$ , which solves to $v_x = \sqrt{\frac{1}{2}Dg}$ . Therefore, the scalar speed of the ball upon entering Portal $2$ in the second frame is $s = \sqrt{v_x^2 + v_y^2} = \sqrt{\sqrt{\frac{1}{2}Dg}^2 + \sqrt{2Dg}^2} = \sqrt{\frac{5}{2}Dg}$ .

In the third frame, where the ball emerges from Portal $1$ and lands past Portal $2$ , this scalar speed is converted to a horizontal velocity, so $v_x = s = \sqrt{\frac{5}{2}Dg}$ . Since the ball will fall the same distance as in the second frame, $t = \sqrt{\frac{2D}{g}}$ once again, and the horizontal distance is $d_x = v_xt = \sqrt{\frac{5}{2}Dg}\sqrt{\frac{2D}{g}} = \sqrt{5}D$ . Therefore, the ball lands $\sqrt{5}D$ from the origin, so $\alpha = \sqrt{5} \approx \boxed{2.23607}$ .