Projectile in Horizontal Projection

Consider a ball thrown horizontally with speed $u$ from the top of a tower of height $h$. It follows a curved path down to the point where it lands (call it A). Because the motion is genuinely two-dimensional, the horizontal ($x$) and vertical ($y$) motions can — and should — be analysed completely separately, using the one-dimensional kinematic equations of §2.10.3 along each axis, and only recombined at the end.

Horizontal motion: there is no horizontal acceleration ($a_x=0$), so the initial horizontal velocity $u$ is preserved throughout, and the horizontal distance covered in time $t$ is

Vertical motion: the initial vertical velocity is zero ($u_y=0$; the launch is purely horizontal), and the vertical acceleration is $g$ downward, so the vertical drop in time $t$ is

The trajectory (path shape). Eliminating $t$ between (2.23) and (2.24) — solving (2.23) for $t=x/u$ and substituting into (2.24) — gives

Since $y \propto x^2$, the path is a parabola — the same conclusion holds for oblique projection in §2.11.3.

(1) Time of flight $T$: the time to reach the ground, $y=h$. From (2.24), $h=\tfrac12gT^2$, so

Notably $T$ depends only on the drop height $h$ and $g$ — it does not depend on the horizontal launch speed at all. A ball simply dropped from height $h$ and a ball launched horizontally from the same height with any speed therefore reach the ground at exactly the same instant.

(2) Horizontal range $R$: the horizontal distance from the foot of the tower to the landing point, $R=uT=u\sqrt{2h/g}$ — directly proportional to the launch speed $u$ and inversely related to $g$ (through the $\sqrt{1/g}$ factor). …