Vectors & Projectile Motion

You’ll be able to

Resolve a vector into perpendicular components and combine components back into a magnitude
Treat 2D motion as two independent 1D problems sharing a clock
Analyze projectile motion using aₓ = 0 and a vertical acceleration of −g

Position, velocity, and acceleration are vectors

In two dimensions the position is a vector r(t) = x(t) î + y(t) ĵ. Differentiating each component gives the velocity vector, and differentiating again gives the acceleration vector. The calculus is identical to 1D — you simply do it once per component, independently.

Component form of the motion vectors

r = x î + y ĵ v = (dx/dt) î + (dy/dt) ĵ a = (dvₓ/dt) î + (dv_y/dt) ĵ

A vector of magnitude V at angle θ resolves as Vₓ = V cos θ and V_y = V sin θ; recombine with |V| = √(Vₓ² + V_y²).

The independence of horizontal and vertical motion

The great simplification of projectile motion: the horizontal and vertical motions do not affect each other — they only share the same clock t. For a projectile the horizontal acceleration is aₓ = 0 (constant horizontal velocity), while the vertical acceleration is −g (a steady 9.8 m/s² downward). Solve the vertical problem for timing, then feed that time into the horizontal problem.

Worked example

A projectile is launched from ground level at v₀ = 25 m/s at θ = 37° above the horizontal. Using g = 10 m/s², cos 37° = 0.8, sin 37° = 0.6, find the time of flight, the horizontal range, and the maximum height.

1.Resolve the launch velocity: horizontal v₀ₓ = 25(0.8) = 20 m/s, vertical v₀y = 25(0.6) = 15 m/s.
2.Vertical velocity is v_y(t) = 15 − 10t. It reaches the top when v_y = 0, at t = 1.5 s; by symmetry the total flight time is twice that, t = 3.0 s.
3.Horizontal motion has no acceleration, so range = v₀ₓ · t = 20 × 3.0 = 60 m.
4.Maximum height uses the vertical motion: H = v₀y²/(2g) = 15²/(2·10) = 225/20 = 11.25 m.

Answer: Time of flight 3.0 s, range 60 m, maximum height 11.25 m

Checkpoint

A particle moves in a plane with r(t) = (3t) î + (4t − 5t²) ĵ (SI units). What is the magnitude of its velocity at t = 0?

Checkpoint

At the highest point of a projectile’s arc (ignoring air resistance), which statement is correct?

Tip

Always split a 2D problem into components before doing anything else. Keep two columns — horizontal (a = 0) and vertical (a = −g) — and remember they are linked only through the shared time t.

Answer the 2 checkpoints as you read.