Projectile Motion

Motion
definition

Also known as: trajectory, ballistic motion

Grade 9-12

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Two-dimensional motion under gravity alone, where horizontal velocity is constant and vertical motion is uniformly accelerated — producing a parabolic path. Projectile motion models the trajectory of thrown balls, kicked footballs, launched rockets, and fired bullets.

Definition

Two-dimensional motion under gravity alone, where horizontal velocity is constant and vertical motion is uniformly accelerated — producing a parabolic path.

💡 Intuition

A thrown ball follows a curved path—horizontal motion is steady, vertical is accelerated.

🎯 Core Idea

Horizontal and vertical motions are independent—analyze separately, then combine.

Example

A ball thrown at 45° travels farthest; at 90° it goes straight up and down.

Formula

x = v_0 \cos\theta \cdot t \quad ; \quad y = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2

Notation

v_0 is the initial speed in m/s, \theta is the launch angle, g \approx 9.81 m/s² is gravitational acceleration, x is horizontal position, y is vertical position, and R is the range.

🌟 Why It Matters

Projectile motion models the trajectory of thrown balls, kicked footballs, launched rockets, and fired bullets. It is central to sports science, military ballistics, space mission planning, and the design of fountains and firework displays.

💭 Hint When Stuck

When solving a projectile problem, split the motion into horizontal (x) and vertical (y) components. Horizontally, velocity is constant: x = v_{0x} t. Vertically, use free-fall equations with a = -g. Find the time of flight from the vertical equation, then use it to find the horizontal range.

Formal View

With launch angle \theta and initial speed v_0, the position is x(t) = v_0 \cos\theta \cdot t and y(t) = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2. The range on level ground is R = \frac{v_0^2 \sin 2\theta}{g}, maximised at \theta = 45°.

🚧 Common Stuck Point

The horizontal velocity stays constant (no horizontal acceleration).

⚠️ Common Mistakes

  • Applying acceleration in the horizontal direction — there is no horizontal acceleration in ideal projectile motion (air resistance is neglected), so horizontal velocity is constant.
  • Using the full initial speed instead of its components — you must resolve v_0 into v_{0x} = v_0\cos\theta and v_{0y} = v_0\sin\theta before applying kinematic equations.
  • Forgetting that at the highest point the vertical velocity is zero but the horizontal velocity is not — the projectile does not stop at the top, it only stops rising.

Frequently Asked Questions

What is Projectile Motion in Physics?

Two-dimensional motion under gravity alone, where horizontal velocity is constant and vertical motion is uniformly accelerated — producing a parabolic path.

What is the Projectile Motion formula?

x = v_0 \cos\theta \cdot t \quad ; \quad y = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2

When do you use Projectile Motion?

When solving a projectile problem, split the motion into horizontal (x) and vertical (y) components. Horizontally, velocity is constant: x = v_{0x} t. Vertically, use free-fall equations with a = -g. Find the time of flight from the vertical equation, then use it to find the horizontal range.

Next Steps

circular motion

How Projectile Motion Connects to Other Ideas

To understand projectile motion, you should first be comfortable with free fall, velocity and vectors. Once you have a solid grasp of projectile motion, you can move on to circular motion.

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