Resolving Vectors to Analyze Projectile Motion

Here is the key insight that unlocks two-dimensional motion: when an object is launched into the air, gravity only pulls it *downward*. It has no effect on how fast it moves *sideways*. That means we can split any velocity vector into two completely separate, independent lanes. This process is called vector decomposition. Start with the launch velocity v₀ at angle θ above horizontal. Using trigonometry: • Horizontal component: v₀ₓ = v₀ cos θ — this stays constant throughout the flight (no horizontal force acts on the projectile). • Vertical component: v₀ᵧ = v₀ sin θ — this changes every second because gravity accelerates the object downward at g ≈ 9.8 m/s². Once decomposed, you solve two separate 1D problems simultaneously. For the vertical lane, use the kinematic equations you already know: y = v₀ᵧ t − ½g t² and v_y = v₀ᵧ − g t. For the horizontal lane, since acceleration is zero: x = v₀ₓ t. A common misconception: many students think the horizontal speed slows down because the projectile is 'fighting gravity.' It does not — horizontal and vertical motions are independent. At maximum height, the vertical velocity is zero, but the projectile is still moving horizontally at full speed. These equations assume no air resistance and that the projectile lands at the same elevation it was launched from. To find the full trajectory under these ideal conditions, solve for time from one equation and substitute into the other. For a projectile landing at the same height it launched from, the time of flight is T = 2v₀ sin θ / g, the maximum height is H = (v₀ sin θ)² / (2g), and the range is R = v₀² sin(2θ) / g. Notice that maximum range occurs at θ = 45° because sin(2 × 45°) = sin 90° = 1, its maximum value.