Lesson 2: Vectors and Projectile Motion

Many quantities in physics have direction, and pretending they do not leads to wrong answers. This lesson builds the two tools for handling direction — resolving vectors into components, and treating horizontal and vertical motion as independent — and combines them in the most analysed motion in all of school physics: the projectile.

Vectors and components

A vector can be fully replaced by two perpendicular components. For a velocity of 25 m/s directed 35° above the horizontal:

vx = 25 cos 35° = 25 × 0.819 = 20 m/s
vy = 25 sin 35° = 25 × 0.574 = 14 m/s

The components add back by Pythagoras: √(20² + 14²) = √(400 + 206) ≈ 25 m/s, as they must. To add two vectors, add their components separately. A force of 3.0 N east and one of 4.0 N north give a resultant √(3² + 4²) = 5.0 N at an angle arctan(4 ÷ 3) = 53° north of east — the famous 3-4-5 triangle. Displacement, velocity, acceleration and force are all vectors; mass, time, energy and temperature are scalars.

Projectile motion

Throw a ball and its path is a parabola. Galileo's insight was that the horizontal and vertical motions are independent: the horizontal velocity stays constant (no horizontal force, ignoring air resistance), while the vertical motion is free fall with acceleration g downwards. At the top of the flight the vertical velocity is zero — but the horizontal velocity is untouched, and the acceleration is still g, pointing down. The ball does not "stop" at the top, and there is no moment of weightlessness.

Worked example 1: full projectile analysis

A ball is projected at 20 m/s at 40° above the horizontal from ground level. Find its maximum height, flight time and range. Resolve: vy = 20 sin 40° = 12.9 m/s; vx = 20 cos 40° = 15.3 m/s. Maximum height (vertical velocity zero at top): H = vy² ÷ 2g = (12.9)² ÷ 19.6 = 165 ÷ 19.6 = 8.4 m. Time to the top = vy ÷ g = 12.9 ÷ 9.8 = 1.31 s, so total flight time = 2 × 1.31 = 2.6 s. Range = vx × flight time = 15.3 × 2.6 = 40 m. For level ground the range is maximised at 45°, and angles equally above and below 45° give the same range — worth remembering when a question offers you two launch angles.

Worked example 2: the cliff projectile

A stone is thrown horizontally at 15 m/s from a 45 m cliff. How far from the cliff base does it land, and how fast is it moving at impact? Vertical motion decides the time: s = ½gt², so 45 = 4.9t², giving t = √(45 ÷ 4.9) = 3.0 s. Horizontal distance = 15 × 3.0 = 45 m (45.5 m before rounding). Vertical velocity at impact = gt = 9.8 × 3.0 = 30 m/s, so impact speed = √(15² + 30²) = √(225 + 882) = 33 m/s, directed at arctan(30 ÷ 15) = 63° below the horizontal. The horizontal launch speed affected only *where* it landed, never *how long* it fell.

In the lab: testing the range prediction

With a spring launcher firing a ball bearing at 45°, suppose your calculation predicts a range of 12.4 m but the mean of ten measured ranges is 11.6 m — 6.5% short. Where does the discrepancy come from? Launch speed from a spring varies slightly shot to shot (±2% is typical), the launch angle setting has ±1° play, and — most importantly — air resistance acts throughout the flight, reducing range, an effect our model ignores. Releasing from a small height above the landing surface works the other way, lengthening the flight. A good lab report names each uncertainty, estimates its size, and states its direction of effect.

Physics in the real world

Long jumpers take off at about 20–22°, not 45°: human legs deliver far less vertical velocity than horizontal, so the optimal angle given the athlete's speed trade-off is much lower. Basketball free throws arc because a steep entry angle makes the hoop effectively "bigger" — the ball can pass through a wider range of positions — at the cost of needing more precise speed control. Firefighters predict where a water jet lands with projectile maths, and artillery has used tables of ranges since before electronic computers; the independence of horizontal and vertical motion is one of the most practically useful ideas in physics.

Common mistakes

• Using the full launch speed in the vertical equations — resolve first, then treat each direction separately.
• Setting acceleration to zero at the top of the flight. Only the vertical *velocity* is zero there.
• Forgetting that flight time is set by the vertical motion alone.
• Using sin where cos belongs: the component *adjacent* to the angle takes cos.
• Calculator in radian mode — check before every trigonometry step.

Exam Experience

1. At the highest point of a projectile's flight (ignoring air resistance), which statement is correct? A) Velocity and acceleration are both zero. B) Velocity is horizontal and acceleration is g downwards. C) Velocity is zero and acceleration is g downwards. D) Acceleration is zero. Answer: B. The vertical component of velocity is momentarily zero, the horizontal component is unchanged, and gravity accelerates the projectile downwards throughout the flight.

2. Two identical balls are released from the same height at the same instant: one is dropped, the other is projected horizontally at 5.0 m/s. Ignoring air resistance: A) the dropped ball lands first. B) the projected ball lands first. C) they land at the same time. D) the projected ball lands later because it travels further. Answer: C. Vertical motion is independent of horizontal motion; both fall the same distance under the same acceleration.

3. A displacement of 6.0 m east and a displacement of 8.0 m north combine to give a resultant of: A) 14 m. B) 10 m at 53° north of east. C) 10 m at 37° north of east. D) 2.0 m. Answer: B. Magnitude √(36 + 64) = 10 m; angle arctan(8 ÷ 6) = 53° from east.

Constructed response (4 marks). A rescue line is launched horizontally at 12 m/s from a cliff 20 m above the sea. (a) Show the line takes about 2.0 s to reach the water. (b) Calculate how far from the cliff base it lands. (c) Calculate the vertical component of its velocity at impact. Scoring guide: (a) 1 mark for t = √(2s ÷ g) = √(40 ÷ 9.8) = 2.0 s. (b) 1 mark for range = 12 × 2.0 = 24 m. (c) 1 mark for vy = gt = 9.8 × 2.0 = 20 m/s (19.8 m/s unrounded). 1 further mark for consistently carrying units and giving answers to 2–3 significant figures. Common error: using 12 m/s in the vertical equation earns no marks for (a) — the launch was horizontal, so initial vertical velocity is zero.

Going further: crossing a river

Relative motion problems are projectile thinking in disguise: two independent velocity components, one resultant. A boat capable of 5.0 m/s in still water must cross a 120 m wide river flowing at 3.0 m/s. Strategy one — aim straight across: the boat's velocity relative to the bank combines 5.0 m/s across with 3.0 m/s downstream. Crossing time = 120 ÷ 5.0 = 24 s, but the current carries the boat 3.0 × 24 = 72 m downstream meanwhile. The resultant speed is √(5.0² + 3.0²) = 5.8 m/s along a diagonal path. Strategy two — aim upstream at angle θ so the downstream drift cancels: the upstream component must equal the current: 5.0 sin θ = 3.0, so sin θ = 0.60 and θ = 37° upstream of straight-across. The across-river component is then 5.0 cos 37° = 4.0 m/s, and the crossing takes 120 ÷ 4.0 = 30 s — six seconds slower, but the boat lands exactly opposite its start. The trade-off is general: the *fastest* crossing ignores drift; the *straightest* crossing takes longer. Pilots solve the identical problem with wind correction angles, and the same vector triangle appears for aircraft velocities, rain slanting on a moving train window, and a swimmer crossing a tidal channel. Whenever two motions combine, draw the vector triangle first — components before conclusions.

In focus: one range, two angles

For projectiles landing at launch height, the range formula R = v² sin 2θ ÷ g (worth deriving once from this lesson's method) reveals a symmetry: sin 2θ takes the same value at θ and at 90° − θ. A ball launched at 20 m/s therefore lands at R = 400 × sin 60° ÷ 9.8 = 35 m whether it is kicked at 30° or at 60°. The flights differ in everything except the landing spot: the 30° launch peaks at (20 sin 30°)² ÷ 2g = 5.1 m and is airborne for 2.0 s; the 60° launch peaks at 15 m and hangs for 3.5 s. Team sports choose between them tactically — a flat, fast pass that arrives quickly versus a high lob that clears defenders but gives them time to move. Artillery historically printed both solutions in its range tables ("low angle" and "high angle" fire for the same target, the latter able to drop shells over hills). The formula has limits: it assumes launch and landing at the same height and negligible air resistance. A real football loses roughly a third of its vacuum range to drag at these speeds, and the optimum angle drops from 45° towards 35–40° — one more reason the clean textbook parabola is a model to reason *from*, not a description to trust blindly.

Formula card

Exam technique: solve the vertical direction first to get the time — the horizontal direction almost never needs anything else. If launch and landing heights differ, the symmetry shortcuts fail; go back to s = ut + ½at² for the vertical motion and everything still works.