Kinematics in Two Dimensions: Projectile Motion

Learning Objectives

Resolve an initial velocity vector into its horizontal (x) and vertical (y) components.

Analyze the horizontal and vertical motion of a projectile independently.

Understand that the horizontal motion has constant velocity and the vertical motion has constant downward acceleration (g).

Solve basic projectile motion problems for time of flight, range, and maximum height.

The Motion of a Projectile

Projectile motion is the motion of an object that is thrown or projected into the air, subject only to the acceleration of gravity. The path the object follows is called its trajectory, which is a parabola.

The Key Principle: Independence of Motion

The crucial insight for solving projectile motion problems is that the horizontal motion and the vertical motion are completely independent of each other. We can analyze them as two separate, one-dimensional kinematics problems.

The only thing that links the two is time (t). The projectile stops moving horizontally at the same time it stops moving vertically (when it hits the ground).

Analyzing the Motion

When an object is launched with an initial velocity (v₀) at an angle (θ) above the horizontal, we first resolve this velocity into its components using trigonometry:

Initial horizontal velocity (v₀x): v₀x = v₀ cos(θ)

Initial vertical velocity (v₀y): v₀y = v₀ sin(θ)

1. Horizontal Motion (x-direction):

There are no forces acting horizontally (we ignore air resistance).

Therefore, the horizontal acceleration is zero (aₓ = 0).

This means the horizontal velocity remains constant throughout the flight (vₓ = v₀x).

The only equation needed is: x = vₓt (distance = speed × time).

2. Vertical Motion (y-direction):

The only force acting vertically is gravity.

Therefore, the vertical acceleration is constant and directed downwards: aᵧ = -g (where g ≈ 9.8 m/s²).

This is a standard free-fall problem. The vertical velocity changes continuously.

We can use the standard kinematic equations:

vᵧ = v₀y + aᵧt

y = v₀yt + ½aᵧt²

Key Quantities

Maximum Height: Occurs at the peak of the trajectory. At this single instant, the vertical velocity (vᵧ) is zero.

Time of Flight: The total time the object is in the air.

Range: The total horizontal distance the object travels.

Example: A cannonball is fired horizontally from a 100m high cliff with a speed of 50 m/s. How far from the base of the cliff does it land?

1.Analyze vertical motion to find time: It falls 100m from rest. y = ½gt². 100 = ½(9.8)t². t² ≈ 20.4. t ≈ 4.5 s.

2.Analyze horizontal motion to find range: The time in the air is 4.5 s. Range x = vₓt = (50 m/s)(4.5 s) = 225 m.

Key Terms

Projectile Motion: The form of motion experienced by an object or particle (a projectile) that is projected near the Earth's surface and moves along a curved path under the action of gravity only.
Trajectory: The path followed by a projectile flying or an object moving under the action of given forces.
Range: The horizontal distance traveled by a projectile.
Vector Components: The parts of a vector that are resolved into perpendicular directions, typically horizontal (x) and vertical (y).