Two Dimensional Kinematics and Projectile Motion

One-dimensional kinematics describes motion along a single straight line, such as a car speeding up on a straight road. Two-dimensional kinematics extends the same ideas to motion in a plane, where an object's position, velocity, and acceleration each have both a horizontal (\(x\)) and a vertical (\(y\)) part. If you have not reviewed the basic motion equations yet, it helps to start with scalars vectors and one dimensional motion before moving into two dimensions.

The most common example of two-dimensional kinematics is projectile motion: the curved path followed by any object launched into the air and acted on only by gravity, such as a thrown ball, a kicked football, or a launched water balloon. Other examples of two-dimensional motion, like horizontal circular motion, follow a curved path for different reasons, but projectile motion is the case you will see most often in an introductory course.

A projectile is launched with an initial speed \(v_0\) at an angle \(\theta\) above the horizontal. Because velocity is a vector, it can be split into two perpendicular components using right-triangle trigonometry:

\(v_{0x} = v_0 \cos\theta\)

\(v_{0y} = v_0 \sin\theta\)

The key idea behind two-dimensional kinematics is that horizontal and vertical motion happen independently and can be analyzed with separate sets of equations that only share one variable: time.

Horizontally, once the projectile leaves the launcher, no horizontal force acts on it (ignoring air resistance), so the horizontal velocity stays constant:

\(x(t) = v_{0x}\,t\)

Vertically, gravity produces a constant downward acceleration \(g \approx 9.8\ \)m/s\(^2\). This is the same acceleration you would find from applying newtons second law of motion to an object acted on only by its weight. The vertical motion follows the familiar constant-acceleration equations:

\(v_y(t) = v_{0y} - g t\)

\(y(t) = v_{0y}\,t - \frac{1}{2} g t^2\)

For a projectile launched from and landing at the same height, three shortcut formulas come directly from the equations above:

Time of flight: \(T = \dfrac{2 v_{0y}}{g}\)

Maximum height: \(H = \dfrac{v_{0y}^2}{2g}\)

Horizontal range: \(R = v_{0x} T = \dfrac{v_0^2 \sin(2\theta)}{g}\)

The range formula shows that, for a fixed launch speed, the range is largest when \(\theta = 45^\circ\), since \(\sin(2\theta)\) reaches its maximum value of 1 at that angle.

A ball is launched at \(v_0 = 20\ \)m/s\(\) at an angle of \(30^\circ\) above level ground. Find the time of flight, maximum height, and range.

Step 1: Find the components.

\(v_{0x} = 20 \cos 30^\circ \approx 17.3\ \)m/s\(\)

\(v_{0y} = 20 \sin 30^\circ = 10\ \)m/s\(\)

Step 2: Time of flight.

\(T = \dfrac{2(10)}{9.8} \approx 2.04\ \)s\(\)

Step 3: Maximum height.

\(H = \dfrac{10^2}{2(9.8)} \approx 5.10\ \)m\(\)

Step 4: Range.

\(R = 17.3 \times 2.04 \approx 35.3\ \)m\(\)

This matches using the shortcut formula directly: \(R = \dfrac{(20)^2 \sin(60^\circ)}{9.8} \approx 35.3\ \)m\(\).

Plotting height against horizontal distance for a projectile always produces a parabola, since \(y\) depends on \(x^2\) once time is eliminated between the horizontal and vertical equations.

Most errors in projectile motion problems come from mixing the two directions together. Keep these points in mind:

Gravity only affects the vertical component, never the horizontal one, so \(v_x\) never changes during flight.

Always confirm whether \(\theta\) is measured from the horizontal or the vertical before using \(\sin\theta\) and \(\cos\theta\).

The range formula \(R = \dfrac{v_0^2 \sin(2\theta)}{g}\) only applies when launch and landing heights are equal; if they differ, you must go back to the full \(x(t)\) and \(y(t)\) equations.

Related lessons

• 2d conservation of momentum