Meeting condition: ( x_1 = x_2 ) ( 3t = 200 - 2t - 0.25 t^2 ) ( 0.25 t^2 + 5t - 200 = 0 ) Multiply by 4: ( t^2 + 20t - 800 = 0 ) ( t = \frac-20 \pm \sqrt400 + 32002 = \frac-20 \pm 602 ) Positive root: ( t = 20 ) seconds.

Stone 2 is thrown 1 second later, so its travel time = t - 1 = 2.193 s. Initial velocity u (downward positive): y = u·t₂ + ½ g t₂² → 50 = u(2.193) + ½ (9.81)(2.193)² ½(9.81)(4.809) = 23.58 Thus 50 = 2.193u + 23.58 → 2.193u = 26.42 → u ≈ 12.04 m/s downward.

Miguel exhaled. It wasn’t just the answer—it was the method . The way Mathalino broke the motion into phases, checking direction changes before integrating absolute values. That was the key he’d missed in lecture.