Classical Mechanics · Chapter 2

Kinematics

The geometry of motion — describing how things move without asking why.

PrerequisitesBasicalgebra⋅Trigonometry(sin,cos)⋅VectorsBasic algebra \cdot Trigonometry (sin, cos) \cdot Vectors
Learning Goals
  • Distinguish position, displacement, velocity, speed, and acceleration.
  • Choose the right constant-acceleration equation based on given and missing variables.
  • Break projectile motion into independent horizontal and vertical components.
  • Derive range, height, and time of flight for ideal projectiles on level ground.
  • Extend 2D projectile reasoning to simple 3D motion with lateral acceleration.

2.1 Motion in One Dimension

Kinematics is the branch of mechanics that describes motion in terms of position, velocity, and acceleration — without reference to the forces that cause that motion. Before Newton can tell us why something accelerates, kinematics gives us the language to describe how it moves.

Definition 2.1 — Average and Instantaneous Velocity
The average velocity overatimeinterval\Deltatisthedisplacementdividedbyelapsedtimeover a time interval \Deltat is the displacement divided by elapsed timevˉ=\Deltax/\Deltatv̄ = \Deltax / \DeltatThe instantaneous velocity isthelimitas\Deltat→0—thederivativeofpositionwithrespecttotimeis the limit as \Deltat \to 0 — the derivative of position with respect to timev(t)=dx/dtv(t) = dx/dtSimilarly, acceleration is the rate of change of velocity: a(t)=dv/dt=d2x/dt2a(t) = dv/dt = d^{2}x/dt^{2}.

For the special case of constant acceleration— which applies to free fall near Earth's surface, and to many idealized problems — we can integrate twice to get the four kinematic equations:

v=v0+atv = v_0 + at(2.1)
x=x0+v0t+12at2x = x_0 + v_0t + \frac{1}{2}at^2(2.2)
v2=v02+2a(x−x0)v^2 = v_0^2 + 2a(x - x_0)(2.3)
x=x0+12(v0+v)tx = x_0 + \frac{1}{2}(v_0 + v)t(2.4)

These four equations contain six variables: x, x₀, v, v₀, a, t. Each equation omits one variable. When solving a problem, identify which variable is unknown and which is not given, then choose the equation that doesn't involve the missing variable.

Example 2.1 — Braking Distance

Acartravelingat30m/s(≈108km/h)brakeswithdeceleration8m/s2.HowfardoesittrA car traveling at 30 m/s (\approx 108 km/h) brakes with deceleration 8 m/s^{2}. How far does it travel before stopping?

Identify:v0=30m/s,v=0m/s,a=−8m/s2.Find\Deltaxv_{0} = 30 m/s, v = 0 m/s, a = -8 m/s^{2}. Find \Deltax.
Choose equation:Sincetimeisnotgivenorneeded,use(2.3):v2=v02+2a\DeltaxSince time is not given or needed, use (2.3): v^{2} = v_{0}^{2} + 2a\Deltax
Solve:0=(30)2+2(−8)\Deltax→\Deltax=900/16=0 = (30)^{2} + 2(-8)\Deltax \to \Deltax = 900/16 = 56.25 m
Sanity check:At 108 km/h, stopping in 56 m sounds right. Highway braking distances are ~50–80 m.

2.2 Two-Dimensional Motion — Projectile

When an object moves in two dimensions under constant gravitational acceleration, we decompose the problem along two independent axes. This is the key insight: the horizontal and vertical components of motion are completely independent of each other.

Theorem 2.1 — Superposition of Motions
In projectile motion (neglecting air resistance), the horizontal velocity componentremainsconstantthroughouttheflight,whiletheverticalcomponentchangesatrate−ent remains constant throughout the flight, while the vertical component changes at rate -vx(t)=v0cos(θ)=constvγ(t)=v0sin(θ)−gtv_{x}(t) = v_{0} cos(\theta) = const \qquad vᵧ(t) = v_{0} sin(\theta) - gtx(t)=v0cos(θ)⋅ty(t)=v0sin(θ)⋅t−12gt2x(t) = v_{0} cos(\theta) \cdot t \qquad y(t) = v_{0} sin(\theta) \cdot t - \frac{1}{2}gt^{2}

From these equations we can derive the trajectory shape. Eliminating t by solving for t from the x-equation and substituting:

y=xtan⁡θ−gx22v02cos⁡2θy = x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta}(2.5)

This is a downward-opening parabola — the trajectory is parabolic. We can also derive closed-form expressions for range R, maximum height H, and time of flight T:

R=v02sin⁡(2θ)gH=v02sin⁡2θ2gT=2v0sin⁡θgR = \frac{v_0^2\sin(2\theta)}{g} \qquad H = \frac{v_0^2\sin^2\theta}{2g} \qquad T = \frac{2v_0\sin\theta}{g}(2.6)

A notable result from (2.6): range is maximized at θ = 45°, and complementary angles (e.g., 30° and 60°) yield the same range. Try this in the simulation below.

Launch angle45°
Initial speed20 m/s
Gravity
Predicted
Range40.8 m
Max height10.2 m
Time of flight2.88 s
Figure 2.1. Interactive projectile simulation. The dashed line shows the predicted parabolic trajectory; the yellow arrow shows the current velocity vector. Try comparing the same launch speed across different gravity presets.

2.3 Projectile Motion in Three Dimensions

In reality, projectiles are not confined to a vertical plane. A crosswind adds a lateral acceleration, turning the flat parabola into a twisted 3D curve. This is why a quarterback must account for wind, and why long-range artillery shells deviate significantly from 2D predictions.

x(t)=v0cos⁡θ t+12axt2y(t)=v0sin⁡θ t−12gt2z(t)=12awindt2x(t) = v_0\cos\theta\,t + \frac{1}{2}a_xt^2 \qquad y(t) = v_0\sin\theta\,t - \frac{1}{2}gt^2 \qquad z(t) = \frac{1}{2}a_\mathrm{wind}t^2(2.7)

With a sideways wind acceleration a_wind, the range projection in the xz-plane is a parabola in both the vertical (xy) and horizontal (xz) senses simultaneously — a spatial parabola.

Figure 2.2. 3D projectile simulation. Set crosswind speed and direction to see the flat parabola bend into 3D space. The ground shadow shows actual landing displacement. Drag to rotate, scroll to zoom.
Example 2.2 — Cliff Launch

A ball is thrown horizontally from a cliff 45 m high with speed 15 m/s. Find where it lands.

Vertical (free fall):y=12gt2→45=12(9.81)t2→t=(90/9.81)=y = \frac{1}{2}gt^{2} \to 45 = \frac{1}{2}(9.81)t^{2} \to t = \sqrt(90/9.81) = 3.03 s
Horizontal:x=v0t=15×3.03=x = v_{0}t = 15 \times 3.03 = 45.5 m from the base of the cliff.
Impact speed:vx=15m/s,vγ=gt=9.81×3.03=29.7m/s→v=(152+29.72)=33.2m/sv_{x} = 15 m/s, vᵧ = gt = 9.81 \times 3.03 = 29.7 m/s \to v = \sqrt(15^{2} + 29.7^{2}) = 33.2 m/s
Definition 2.2 — Common Traps
  • Distance is not displacement: displacement includes direction and can be zero after a round trip.
  • Velocity is not speed: velocity can be negative; speed is the magnitude.
  • Use signs consistently: ifupispositive,gravitationalaccelerationis−gif up is positive, gravitational acceleration is -g
  • Horizontal and vertical motion share time only: do not use horizontal velocity in vertical equations.
  • Range formulas have assumptions: the closed forms above require level launch and landing heights with no air resistance.
Exercises — 2.1–2.3 Kinematics
1.
Aballislaunchedat45°withv0=25m/sonflatground.FindtheA ball is launched at 45° with v_{0} = 25 m/s on flat ground. Find the range. (g=9.81m/s2g = 9.81 m/s^{2}
m
Straightforward
2.
A ball is thrown straight upward and reaches a maximum height of 20 m. What was the initial upward velocity?
m/s
Straightforward
3.
A stone is thrown horizontally from the top of a building 80 m tall. What minimum horizontal speed is needed to land 120 m away?
m/s
Intermediate
4.
A targetis50mawayonflatground.Whatistheminimumlaunchspeedneededtoreachit?(g=get is 50 m away on flat ground. What is the minimum launch speed needed to reach it? (g = m/s2m/s^{2}
m/s
Intermediate
5.Provealgebraicallythatcomplementarylaunchangles(θand90°−θ)alwaysproducethesaProve algebraically that complementary launch angles (\theta and 90° - \theta) always produce the same range on flat ground. Do they also reach the same maximum height?
Challenging
Key Takeaways
  • The four kinematic equations apply whenever acceleration is constant.
  • In projectile motion, horizontal and vertical components are independent — solve each axis separately.
  • Rangeismaximizedatθ=45°;complementaryanglesgiveidenticalrangeRange is maximized at \theta = 45°; complementary angles give identical range.
  • Trajectory shape is parabolic (quadratic in x).
  • OntheMoon(g=1.62m/s2),thesamelaunchproducesa6×longerrangethanonEarthOn the Moon (g = 1.62 m/s^{2}), the same launch produces a 6\times longer range than on Earth.