5.3 Projectile Motion

Projectile Motion

Projectile motion describes what happens when an object is launched into the air and moves under the influence of gravity alone. Understanding it requires treating horizontal and vertical motion as two separate, independent problems that share one common link: time. This concept shows up constantly in physics, from analyzing a basketball's arc to predicting where a launched rocket stage will land.

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Properties of Projectile Motion

A projectile is any object launched into the air that, after launch, moves only under the influence of gravity (no engine, no thrust). The curved path it follows is called its trajectory, and without air resistance, that trajectory is always a parabola.

The central idea is that horizontal and vertical motion are independent of each other. You can analyze them separately:

• Horizontal motion: No force acts horizontally (ignoring air resistance), so horizontal velocity stays constant the entire flight. There is zero horizontal acceleration.
• Vertical motion: Gravity pulls the object downward at a constant acceleration of $g = 9.81 \, m/s^2$. Vertical velocity changes throughout the flight, decreasing on the way up, reaching zero at the peak, and increasing on the way down.

What ties these two components together is time. At any given moment, the projectile has both a horizontal position and a vertical position. The clock runs the same for both directions.

The projectile's trajectory is fully determined by two things: the magnitude of the initial velocity (how fast it's launched) and the direction (the launch angle relative to the horizontal).

Calculations with Kinematic Equations

To solve projectile problems, start by breaking the initial velocity into its two components:

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

where $v_0$ is the launch speed and $\theta$ is the launch angle measured from the horizontal.

From there, use the standard kinematic equations applied to each direction separately.

Horizontal equations (constant velocity, no acceleration):

• Position: $x = v_{0x} \, t = v_0 \cos(\theta) \, t$
• Velocity: $v_x = v_{0x}$ (stays the same throughout)

Vertical equations (constant acceleration $g$ downward):

• Position: $y = v_{0y} \, t - \frac{1}{2}g t^2 = v_0 \sin(\theta) \, t - \frac{1}{2}g t^2$
• Velocity: $v_y = v_{0y} - g t = v_0 \sin(\theta) - g t$

Solving a typical problem step by step:

1. Identify the known quantities: initial speed, launch angle, and what you're solving for (time, range, max height, etc.).
2. Resolve $v_0$ into $v_{0x}$ and $v_{0y}$ using sine and cosine.
3. Pick the appropriate equation for the direction that contains your unknown.
4. Solve for the unknown. Often you'll need to find time from the vertical equations first, then plug that time into the horizontal equation.

Example: A ball is kicked at $20 \, m/s$ at $30°$ above the horizontal. Find how far it travels before hitting the ground (same launch height).

1. $v_{0x} = 20 \cos(30°) = 17.3 \, m/s$

2. $v_{0y} = 20 \sin(30°) = 10.0 \, m/s$

3. Find total flight time using $y = v_{0y} t - \frac{1}{2}g t^2$. Setting $y = 0$ (returns to launch height): $0 = 10.0 \, t - \frac{1}{2}(9.81) t^2$. Factor out $t$: $t(10.0 - 4.905 \, t) = 0$, giving $t = 0$ (launch) or $t = 2.04 \, s$.

4. Horizontal range: $x = 17.3 \times 2.04 = 35.3 \, m$.

Range and Maximum Height

For a projectile launched and landing at the same height, there are shortcut formulas worth knowing.

Range (total horizontal distance):

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

Because $\sin(2\theta)$ reaches its maximum value of 1 when $2\theta = 90°$, the maximum range occurs at a launch angle of 45°. Also note that complementary angles (like 30° and 60°) produce the same range, since $\sin(2 \times 30°) = \sin(2 \times 60°) = \sin(60°) = \sin(120°)$.

Maximum height (peak of the trajectory):

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

The projectile reaches its peak when vertical velocity equals zero. The time to reach that peak is:

$t_{max} = \frac{v_0 \sin(\theta)}{g}$

This is exactly half the total flight time (for same-height launch and landing), which reflects the symmetry of projectile motion: the ascending path mirrors the descending path. The speed at any height on the way up equals the speed at that same height on the way down.

Fundamental Principles and Applications

Projectile motion is a direct consequence of Newton's laws. After launch, the only force acting on the projectile is gravity (in the ideal case), which produces a constant downward acceleration. No horizontal force means no horizontal acceleration, which is why horizontal velocity stays constant.

In the real world, air resistance causes deviations from the ideal parabolic path. Drag reduces both range and maximum height, and the trajectory becomes asymmetric, with a steeper descent than ascent. The optimal launch angle for maximum range shifts below 45° when air resistance is significant.

The study of projectile motion, called ballistics, applies across many fields:

• Sports: Analyzing the trajectory of a soccer ball, golf ball, or basketball shot
• Engineering: Designing water fountains, irrigation sprinklers, or launch systems
• Forensics and military science: Determining bullet trajectories or artillery targeting