Horizontal Projection — Time of Flight

The situation

A body is projected horizontally — at angle $\theta = 0°$ — from a height $h$ above the ground. It has initial horizontal velocity $u$ and zero initial vertical velocity.

It travels forward while simultaneously falling under gravity. The question: how long before it hits the ground?

Key observation

Since the body is launched horizontally, it has no initial vertical velocity. The vertical motion is identical to a body simply dropped from height $h$ — the horizontal motion has no effect on how fast it falls.

Derivation

Take downward as positive for the vertical direction. The body starts at height $h$ above the ground, so it must fall a vertical distance $h$.

Initial vertical velocity: $u_y = 0$ (horizontal launch)

Vertical acceleration: $+g$ (downward)

Using the second equation of motion for vertical displacement:

$h = u_y \cdot T + \frac{1}{2}gT^2$

$h = 0 \cdot T + \frac{1}{2}gT^2$

$h = \frac{1}{2}gT^2$

$T^2 = \frac{2h}{g}$

$\boxed{T = \sqrt{\frac{2h}{g}}}$

What this tells us

The time of flight depends only on the height $h$ and gravity $g$. It does not depend on the horizontal velocity $u$ at all.

A ball rolled off a table at $2$ m/s and one rolled off at $20$ m/s both hit the ground at exactly the same time — provided the table is the same height.

This is a direct consequence of the independence of horizontal and vertical motions.

Comparison with free fall

A body dropped from rest from height $h$ also takes time $T = \sqrt{\frac{2h}{g}}$ to reach the ground. The horizontally projected body takes exactly the same time — it just lands further away.

This is one of the most striking results in kinematics.