Read and reason
A surprisingly equal fall time
Imagine two small objects at the same height: one is released with no initial horizontal velocity; the other is launched horizontally. In an ideal model with negligible air resistance, their vertical positions match at each time because both start with the same vertical velocity and experience the same gravitational acceleration. Horizontal motion does not make one fall more slowly. This conceptual comparison does not require actually dropping or launching anything.
Choose axes and declare assumptions
Choose +x right and +y upward. Near Earth's surface, approximate acceleration as ax=0 m/s² and ay=−9.8 m/s². The initial height y0 and vertical velocity vy0 determine the vertical motion: y(t)=y0+vy0t−4.9t². Horizontal motion is x(t)=x0+vx0t when horizontal acceleration is zero. Time t is shared between both equations, so the components are independent in their equations but still describe one trajectory.
Why impact time depends on vertical motion
A horizontal launch has vy0=0, so for level ground impact y(t)=0 gives 0=y0−4.9t². For y0=19.6 m, t²=4 and t=2 s. Both a dropped object and a horizontally launched object from that height reach ideal level ground in 2 s. The horizontally launched object meanwhile has a nonzero x displacement. A different initial vertical speed or a different landing height changes the impact time.
Cautions about the idealization
Air resistance, uneven terrain, differing release times, wind and object shapes can spoil the ideal equality. Earth curvature and variation in gravity are ignored. If objects are released at different heights or with different vertical velocities, they do not share impact time. A good conclusion identifies the chosen vertical initial conditions, the assumption of negligible drag, and the physical boundary being compared.