Projectile Motion: Why Horizontal Speed Doesn't Change Fall Time

A bowling ball and a cannonball leave the wall top at the same instant: one dropped, one fired horizontally at v₀. Mass never enters — every ball falls with the same g. Both start with v_y = 0 and both feel the same g, so their heights agree at every instant — the rungs between the ghost images stay level — and they land together after t_fall = √(2h/g). v₀ appears nowhere in that expression: throw at 1 m/s or 100 m/s and the fall takes the same time.

Horizontal and vertical motion are independent. After launch nothing pushes the ball sideways, so v_x stays at v₀ and the horizontal gaps between ghosts are equal; gravity acts only downward, so v_y grows by 9.8 m/s every second, exactly as for the dropped ball, and the vertical gaps grow. The throw decides only how far sideways the ball gets in that fixed time: Δx = v₀·t_fall. To hit the dragon, choose v₀ = d / t_fall — t_fall is in the box above the scene, and a landing within 1.8 m of the dragon's centre counts.

Model: no air resistance, level ground. With drag the thrown ball, moving faster through the air, meets a larger resistive force whose vertical part slows its fall slightly, so it lands a little after the dropped one — the small print behind Galileo's “always”.