Kinematics (1D & 2D motion, projectiles)
The idea
Before you can explain why something moves, you need a precise way to describe how it moves — that is kinematics. You already work with speed, velocity, and acceleration; kinematics turns those ideas into a toolkit of equations that connect position, velocity, acceleration, and time whenever acceleration is constant: v = v₀ + at, Δx = v₀t + ½at², and v² = v₀² + 2aΔx. Pick the equation that contains the three quantities you know and the one you want.
The mental model for two-dimensional motion is independence: horizontal and vertical motion run on separate clocks that happen to read the same time. A projectile's horizontal velocity stays constant (no horizontal force, ignoring air resistance) while its vertical velocity changes by 9.8 m/s every second because of gravity. The two motions share only the time variable, so you solve one direction to find t and feed it into the other.
The classic misconception is that an object moving horizontally falls more slowly than one dropped straight down. It does not: a ball rolled off a table and a ball dropped from table height hit the floor at the same instant, because gravity acts on the vertical motion alone. Horizontal speed never buys a projectile extra hang time.
Worked example
A ball rolls off a horizontal cliff edge 19.6 m above the ground, leaving the edge with a horizontal speed of 12 m/s. How long is it in the air, how far from the base of the cliff does it land, and how fast is it moving on impact? Ignore air resistance.
- Split the motion into independent parts: horizontally the velocity stays 12 m/s the whole flight; vertically the ball starts with zero velocity and accelerates downward at g = 9.8 m/s².
- Use the vertical motion to find the time: the drop obeys h = ½gt², so 19.6 = ½ × 9.8 × t², giving t² = 19.6/4.9 = 4.0 and t = 2.0 s.
- Feed that time into the horizontal motion: range = horizontal speed × time = 12 × 2.0 = 24 m from the base of the cliff.
- For impact speed, find the final vertical velocity first: it grows from zero at 9.8 m/s each second, so after 2.0 s it is 9.8 × 2.0 = 19.6 m/s downward.
- Combine the two perpendicular velocity components with the Pythagorean theorem: speed = √(12² + 19.6²) = √(144 + 384.16) = √528.16 ≈ 23 m/s.
- Sanity-check: the time depended only on the 19.6 m height, not on the 12 m/s launch speed — a dropped ball would land at the same moment, just at the cliff's base.
Answer. The ball is airborne for 2.0 s, lands 24 m from the cliff base, and strikes the ground at about 23 m/s.
Check your understanding
- Why does the horizontal launch speed have no effect on how long a projectile stays in the air?
- How do you decide which constant-acceleration equation to use when a problem gives you three known quantities?
- What changes in the analysis if the ball is launched at an angle above the horizontal instead of rolling off horizontally?
- Where does the independence of horizontal and vertical motion break down for a real object like a badminton shuttle?
Build the foundations first
Kinematics (1D & 2D motion, projectiles) builds on these concepts. If any feel shaky, start there.