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Physics: Motion to AP Physics 1 · Module 2 of 8

Vectors & Projectile Motion

Breaking vectors into components, and treating horizontal and vertical motion independently to solve projectile problems.

In this module
  1. Vectors & Components
  2. Horizontally Launched Projectiles
  3. Projectiles Launched at an Angle
  4. Check yourself

Vectors & Components

  • A vector has magnitude and direction. Resolve it into perpendicular components: vₓ = v cos θ and v_y = v sin θ (θ measured from the horizontal).
  • Add vectors by adding their components, then recombine: magnitude = √(vₓ² + v_y²), direction θ = tan⁻¹(v_y / vₓ).
Worked example

A plane flies at 100 m/s at 30° above the horizontal. Find its horizontal and vertical velocity components.

Solution. vₓ = 100 cos 30° ≈ 86.6 m/s and v_y = 100 sin 30° = 50 m/s.

Horizontally Launched Projectiles

  • Horizontal and vertical motions are independent. Horizontally the velocity is constant (no acceleration); vertically the object is in free fall.
  • Time in the air is set by the vertical motion alone: for a launch from height h with no vertical speed, t = √(2h/g).
Worked example

A ball rolls off a 1.25 m high table at 2.0 m/s. How far from the table does it land?

Solution. t = √(2 × 1.25 / 9.8) ≈ 0.51 s. Horizontal distance = (2.0 m/s)(0.51 s) ≈ 1.0 m.

Common mistake: A ball launched horizontally and a ball dropped from the same height hit the ground at the same time. Horizontal speed doesn't change fall time.

Projectiles Launched at an Angle

  • Split the launch velocity into components. v_y decreases by 9.8 m/s every second; vₓ stays constant the whole flight.
  • At the peak, v_y = 0 and the speed equals vₓ. On level ground the flight is symmetric: time up = time down, and range R = v² sin(2θ)/g, which is greatest at 45°.
Projectile launched at 30°
x (m)y (m)v₀ = 20 m/s at 30°peak: v_y = 0vₓ = 17.3 m/sR ≈ 35 m
Launched at 20 m/s and 30°: the horizontal velocity (17.3 m/s) stays constant, the vertical velocity is zero at the peak (about 5.1 m high), and the ball lands about 35 m away after about 2.04 s.
Worked example

A ball is kicked at 20 m/s at 30° above level ground. Find its time of flight and range.

Solution. vₓ = 20 cos 30° ≈ 17.3 m/s, v_y = 10 m/s. Time up = 10/9.8 ≈ 1.02 s, so flight time ≈ 2.04 s. Range ≈ 17.3 × 2.04 ≈ 35 m (check: v² sin 60°/g = 400 × 0.866 / 9.8 ≈ 35 m).

Check yourself

2 of this module's 5 quiz questions. Pick your answers, then check them in the app, with an explanation for every question.

1. During a projectile's flight (no air resistance), which quantity stays constant?

2. At the highest point of a projectile's path, its vertical velocity is:

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