3.2 Projectile Motion

Projectile Motion Fundamentals

Projectile motion is what happens when an object moves through the air under the influence of gravity alone (ignoring air resistance). It combines two simpler motions: constant-speed horizontal motion and uniformly accelerated vertical motion. Understanding how these two components work independently is the key to solving every projectile problem you'll encounter.

Motion of Projectiles

The core idea behind projectile motion is that you can split it into two separate problems: horizontal (x) and vertical (y). These two components are completely independent of each other.

• Horizontal motion has no acceleration (ignoring air resistance), so the object moves at a constant velocity in the x-direction.
• Vertical motion is governed by gravitational acceleration ($g = 9.8 \, \text{m/s}^2$ downward), so the object speeds up or slows down vertically just like in free fall.

Horizontal launch means the object starts with zero initial vertical velocity. Think of a ball rolling off the edge of a table: it moves forward at a steady pace while falling faster and faster. This produces a curved, parabolic path.

Angular launch means the object is launched at some angle $\theta$ above the horizontal, giving it both a horizontal and a vertical component of initial velocity. A javelin throw or a kicked soccer ball are good examples.

To find those components, you resolve the initial velocity $v_0$ using trigonometry:

• Horizontal component: $v_{0x} = v_0 \cos(\theta)$
• Vertical component: $v_{0y} = v_0 \sin(\theta)$

These two values become the starting points for your separate x and y calculations.

Projectile Trajectory Calculations

Three quantities come up constantly in projectile problems: range, time of flight, and maximum height. Each one follows from applying kinematic equations to the x and y components separately.

Time of flight is the total time the projectile stays in the air. For an object launched from and landing at the same height:

$t = \frac{2v_0 \sin(\theta)}{g}$

This comes from finding how long it takes for the vertical velocity to go up, reach zero, and come back down. For example, a ball launched at 20 m/s at 30° has a flight time of $t = \frac{2(20)\sin(30°)}{9.8} \approx 2.04 \, \text{s}$.

Range is the total horizontal distance traveled during that time:

$R = v_0 \cos(\theta) \cdot t$

For a symmetric launch (same launch and landing height), you can combine this with the time of flight formula to get:

$R = \frac{v_0^2 \sin(2\theta)}{g}$

Notice that $\sin(2\theta)$ is maximized when $2\theta = 90°$, meaning $\theta = 45°$ gives the maximum range on level ground.

Maximum height is the peak of the trajectory, where the vertical velocity momentarily equals zero:

$h = \frac{v_0^2 \sin^2(\theta)}{2g}$

Advanced Projectile Concepts

Finding Initial Velocity for a Target

Sometimes you know where the projectile needs to land and need to figure out how fast to launch it. This is a reverse calculation.

1. Identify the target's coordinates $(x, y)$ relative to the launch point.

2. Write the horizontal equation: $x = v_0 \cos(\theta) \cdot t$.

3. Write the vertical equation: $y = v_0 \sin(\theta) \cdot t - \frac{1}{2}g t^2$.

4. Solve one equation for $t$ and substitute into the other. This often produces a quadratic equation you'll need to solve.

5. If the problem asks for the minimum initial speed to reach a target, you may need to optimize with respect to $\theta$, which is a calculus-based technique that goes beyond most intro courses.

Independence of Horizontal and Vertical Motion

This principle is worth emphasizing because it's the foundation of every projectile calculation.

• The horizontal and vertical components of motion do not influence each other. A bullet fired horizontally from a gun and a bullet dropped from the same height at the same instant will hit the ground at the same time. The horizontal velocity doesn't affect how fast something falls.
• In the x-direction, there's no acceleration, so $x = v_{0x} \cdot t$.
• In the y-direction, gravity accelerates the object, so $y = v_{0y} \cdot t - \frac{1}{2}g t^2$.
• The actual path you observe (the parabola) is the superposition of these two independent motions. You reconstruct the full trajectory by combining the x and y positions at each moment in time.