KAHIBARO
Table of Contents
Motion after a Horizontal Launch
Horizontal projection is a special case of projectile motion in which an object is launched horizontally, not upward or downward. This means its initial velocity has only a horizontal component. The vertical component of the initial velocity is zero.
A common example is a ball rolling off the edge of a table. At the instant it leaves the table, it is moving sideways, but gravity immediately begins pulling it downward. As a result, the object moves forward while also falling.
Initial Conditions
To describe horizontal projection, it is helpful to choose axes so that the $x$ direction is horizontal and the $y$ direction is vertical. If upward is taken as positive $y$, then for an object launched horizontally from height $h$ with speed $v_0$,
$$v_{0x} = v_0, \qquad v_{0y} = 0$$
The horizontal acceleration is zero if air resistance is neglected, and the vertical acceleration is due to gravity:
$$a_x = 0, \qquad a_y = -g$$
For horizontal projection,
$$v_{0y} = 0$$
but
$$a_y = -g \neq 0$$
The object does not start with vertical velocity, but it does have vertical acceleration from the very beginning.
Horizontal and Vertical Motion
The key idea is that the horizontal motion and vertical motion can be treated separately.
Horizontally, there is no acceleration, so the horizontal speed remains constant:
$$x = v_0 t$$
Vertically, the object falls exactly like an object dropped from rest:
$$y = h - \frac{1}{2}gt^2$$
if the object starts at height $h$ above the ground and upward is positive.
This separation is one of the most important features of projectile motion. The forward motion does not affect the falling motion, and the falling motion does not affect the constant horizontal motion.
Time of Fall
Since the object starts with no vertical velocity, the time it takes to hit the ground depends only on the vertical motion.
If the ground is at $y=0$, then
$$0 = h - \frac{1}{2}gt^2$$
Solving for $t$ gives
$$t = \sqrt{\frac{2h}{g}}$$
This result shows that the time to fall depends only on the launch height and gravity, not on the horizontal launch speed.
For a horizontal launch from height $h$,
$$t = \sqrt{\frac{2h}{g}}$$
The fall time is independent of the horizontal speed.
Horizontal Range for a Horizontal Launch
The horizontal distance traveled before hitting the ground is found from the constant horizontal motion:
$$R = v_0 t$$
Substituting the fall time,
$$R = v_0 \sqrt{\frac{2h}{g}}$$
This gives the range for horizontal projection.
A faster horizontal launch gives a larger range, but it does not change how long the object stays in the air.
Velocity During the Motion
As the object moves, its horizontal velocity remains constant:
$$v_x = v_0$$
Its vertical velocity changes because of gravity:
$$v_y = -gt$$
The total velocity is the combination of these two components. Its magnitude is
$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{v_0^2 + g^2 t^2}$$
So the speed increases during the motion because the downward component becomes larger and larger.
Shape of the Path
The path of a horizontally projected object is curved. To see this, eliminate time from the equations.
From
$$x = v_0 t$$
we get
$$t = \frac{x}{v_0}$$
Substitute this into the vertical equation:
$$y = h - \frac{1}{2}g\left(\frac{x}{v_0}\right)^2$$
So the trajectory is
$$y = h - \frac{g}{2v_0^2}x^2$$
This is the equation of a parabola.
The trajectory of a horizontally projected object is parabolic:
$$y = h - \frac{g}{2v_0^2}x^2$$
Comparison with Free Fall
A useful comparison is between two objects released from the same height at the same time. One is simply dropped, and the other is launched horizontally. Both have the same vertical motion because both start with
$$v_{0y} = 0$$
and both have vertical acceleration
$$a_y = -g$$
So they hit the ground at the same time, assuming air resistance is negligible.
This is an important physical idea. Horizontal motion does not delay falling.
Example Setup
Suppose a ball leaves a table of height $1.25 \,\text{m}$ with horizontal speed $3.0 \,\text{m/s}$.
The time of fall is
$$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(1.25)}{9.8}} \approx 0.51 \,\text{s}$$
The horizontal distance traveled is
$$R = v_0 t = 3.0 \times 0.51 \approx 1.53 \,\text{m}$$
So the ball lands about $1.5 \,\text{m}$ from the table.
Summary Table
| Quantity | Horizontal direction | Vertical direction |
|---|---|---|
| Initial velocity | $v_{0x} = v_0$ | $v_{0y} = 0$ |
| Acceleration | $a_x = 0$ | $a_y = -g$ |
| Position equation | $x = v_0 t$ | $y = h - \frac{1}{2}gt^2$ |
| Velocity equation | $v_x = v_0$ | $v_y = -gt$ |
Simple Diagram
Final Idea
Horizontal projection is the simplest two dimensional projectile motion case. The object moves with constant speed sideways while accelerating downward under gravity. The most important fact is that the horizontal and vertical motions are independent, which allows the motion to be analyzed using simple equations for each direction.