Projectile Motion Equations with Detailed Derivations

Consider a projectile launched with initial velocity \(v_0\) at angle \(\theta\) above the horizontal. Neglecting air resistance, the only force is gravity with acceleration \(g = 9.8 \text{ m/s}^2\).

Explore projectile motion interactively: Interactive Simulation. Practice with solved examples: Projectile Problems with Solutions.

Projectile motion trajectory diagram — Projectile motion components and trajectory

Velocity, Acceleration, and Displacement

The initial velocity vector \(\vec{V}_0\) has components:

\[ V_{0x} = V_0 \cos\theta \quad \text{and} \quad V_{0y} = V_0 \sin\theta \]

Acceleration components (constant):

\[ a_x = 0 \quad \text{and} \quad a_y = -g = -9.8 \text{ m/s}^2 \]

Velocity Components Over Time

Horizontal velocity remains constant:

\[ V_x(t) = V_0 \cos\theta \]

Vertical velocity changes uniformly:

\[ V_y(t) = V_0 \sin\theta - g t \]

Displacement Components

Horizontal displacement:

\[ x(t) = V_0 \cos\theta \cdot t \]

Vertical displacement:

\[ y(t) = V_0 \sin\theta \cdot t - \frac{1}{2} g t^2 \]

Trajectory Equation (Parabolic Path)

Eliminate time \(t\) from displacement equations:

From \(x(t)\): \(t = \frac{x}{V_0 \cos\theta}\)

Substitute into \(y(t)\):

\[ y = \frac{V_0 \sin\theta \cdot x}{V_0 \cos\theta} - \frac{1}{2} g \left( \frac{x}{V_0 \cos\theta} \right)^2 \]

Simplify to obtain the parabolic trajectory:

\[ y = x \tan\theta - \frac{g x^2}{2 V_0^2 \cos^2\theta} \]

This has the form \(y = Ax^2 + Bx\) where:

\[ A = -\frac{g}{2 V_0^2 \cos^2\theta} \quad \text{and} \quad B = \tan\theta \]

Time of Flight

Total time projectile remains airborne: solve \(y(t) = 0\) for \(t > 0\)

\[ V_0 \sin\theta \cdot t - \frac{1}{2} g t^2 = 0 \] \[ t\left(V_0 \sin\theta - \frac{1}{2} g t\right) = 0 \]

Non-zero solution:

\[ T = \frac{2 V_0 \sin\theta}{g} \]

Maximum Height

At peak height (\(B\)), vertical velocity \(V_y = 0\):

\[ V_0 \sin\theta - g t = 0 \quad \Rightarrow \quad t = \frac{V_0 \sin\theta}{g} \]

Substitute into \(y(t)\):

\[ H = V_0 \sin\theta \left( \frac{V_0 \sin\theta}{g} \right) - \frac{1}{2} g \left( \frac{V_0 \sin\theta}{g} \right)^2 \] \[ H = \frac{V_0^2 \sin^2\theta}{2g} \]

Horizontal Range

Total horizontal distance when projectile lands (\(C\)):

\[ R = x(T) = V_0 \cos\theta \cdot T = V_0 \cos\theta \cdot \frac{2 V_0 \sin\theta}{g} \] \[ R = \frac{V_0^2 \sin(2\theta)}{g} \]