2.1.4.2. Oblique Projection

Table of Contents

Launch at an Angle

Oblique projection is projectile motion in which an object is launched with an initial speed at some angle above or below the horizontal. Unlike horizontal projection, the object now has both a horizontal and a vertical initial velocity. The motion is still treated as two independent parts, horizontal motion and vertical motion, but now both parts begin with nonzero values.

If the object is launched with initial speed $v_0$ at an angle $\theta$ above the horizontal, then the initial velocity can be split into components:
$$
v_{0x} = v_0 \cos\theta
$$
and
$$
v_{0y} = v_0 \sin\theta
$$

These two components completely describe the launch direction and speed.

For oblique projection near Earth's surface, neglecting air resistance,
$$
a_x = 0, \qquad a_y = -g
$$
The horizontal velocity stays constant, while the vertical velocity changes because of gravity.

Horizontal and Vertical Motion Together

The horizontal and vertical motions happen at the same time, but they are calculated separately.

In the horizontal direction, there is no acceleration, so:
$$
x = x_0 + v_{0x} t = x_0 + v_0 \cos\theta \, t
$$

In the vertical direction, the motion has constant downward acceleration:
$$
y = y_0 + v_{0y} t - \frac{1}{2}gt^2
$$
so
$$
y = y_0 + v_0 \sin\theta \, t - \frac{1}{2}gt^2
$$

The velocity components at time $t$ are:
$$
v_x = v_0 \cos\theta
$$
and
$$
v_y = v_0 \sin\theta - gt
$$

This means the projectile moves forward at constant horizontal speed while rising and then falling vertically.

Shape of the Path

Because the horizontal motion is uniform and the vertical motion is accelerated, the path of the projectile is curved. In fact, the trajectory is a parabola.

A launch at a small angle gives a flatter path. A launch at a larger angle gives a steeper path. At $\theta = 45^\circ$, many projectile situations give an especially large range when launch and landing heights are the same, though the full discussion of range belongs to a separate topic.

Rendering scientific sketch…

Oblique projection of a projectile

Using Components

Oblique projection problems are usually solved by following a clear pattern. First, resolve the initial velocity into horizontal and vertical components. Then write the horizontal equation and the vertical equation separately. Finally, connect them through the common time $t$.

The table below summarizes the main quantities.

Quantity	Horizontal direction	Vertical direction
Initial velocity	$v_{0x} = v_0\cos\theta$	$v_{0y} = v_0\sin\theta$
Acceleration	$a_x = 0$	$a_y = -g$
Velocity at time $t$	$v_x = v_0\cos\theta$	$v_y = v_0\sin\theta - gt$
Position at time $t$	$x = x_0 + v_0\cos\theta \, t$	$y = y_0 + v_0\sin\theta \, t - \frac{1}{2}gt^2$

The angle $\theta$ is measured from the horizontal unless stated otherwise. Then,
$$
v_{0x} = v_0\cos\theta, \qquad v_{0y} = v_0\sin\theta
$$
If the angle is measured from the vertical, the component formulas change.

Physical Interpretation

At the moment of launch, the projectile has an upward part of motion and a forward part of motion. Gravity does not affect the forward motion directly, but it constantly reduces the upward vertical velocity. Eventually the vertical velocity becomes zero at the highest point. After that, the vertical velocity becomes negative, meaning the object moves downward.

The horizontal velocity remains unchanged throughout the flight if air resistance is neglected. This is one of the most important ideas in oblique projection.

Common Special Cases

If $\theta = 0^\circ$, then
$$
v_{0y} = 0
$$
and the motion becomes horizontal projection.

If $\theta = 90^\circ$, then
$$
v_{0x} = 0
$$
and the motion becomes purely vertical.

For angles between these values, the projectile has both horizontal and vertical motion.

A Simple Example

Suppose a ball is thrown with speed $20 \,\text{m/s}$ at an angle of $30^\circ$ above the horizontal. Then
$$
v_{0x} = 20\cos 30^\circ \approx 17.3 \,\text{m/s}
$$
and
$$
v_{0y} = 20\sin 30^\circ = 10 \,\text{m/s}
$$

So the motion equations are
$$
x = 17.3t
$$
and
$$
y = 10t - \frac{1}{2}gt^2
$$

If we take $g = 9.8 \,\text{m/s}^2$, then
$$
y = 10t - 4.9t^2
$$

This pair of equations describes the full motion.

Key Idea to Remember

Oblique projection is best understood as the combination of two simpler motions, constant velocity horizontally and constant acceleration vertically. The launch angle determines how much of the initial speed goes into each direction.

To solve oblique projection,
$$
\text{split } v_0 \text{ into components, then treat } x \text{ and } y \text{ separately}
$$
The link between the two motions is the same time $t$.

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