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Module 2: Motion in Two and Three Dimensions

 

PHYS-2325 M2L2b Projectile Motion, Circular Motion & Relative Motion

 DRAFT — Pending CME Approval

This lesson page is a proposed draft created for structural review. The original M2L2 content has been split into two lessons (M2L2a and M2L2b) to allow appropriate depth for each topic area. This split is pending approval from the Course Matter Expert. Learning objectives, video titles, and final content may be revised before publication.

"In my studies of astronomy and philosophy I hold this opinion about the universe: that the Sun remains fixed in the centre of the circle of heavenly bodies, without changing its place; and the Earth, turning upon itself, moves round the Sun."
— Galileo Galilei


With the vector kinematics framework from M2L2a in hand, you are ready to analyze three of the most important and practical types of 2D motion. Projectile motion applies when only gravity acts on an object — the horizontal velocity is constant while vertical motion is uniformly accelerated, giving the characteristic parabolic trajectory. Uniform circular motion reveals a surprising truth: moving in a circle at constant speed still requires a constant centripetal acceleration directed toward the center. Relative motion shows how the velocity and acceleration of an object depend on who is doing the measuring — a concept essential for analyzing collisions, weather systems, and spacecraft maneuvers. All five sample problem videos from your course are in this lesson.

A composite image showing a parabolic projectile path, a circular orbit with centripetal acceleration arrow, and two reference frames illustrating relative motion.
Three distinct types of 2D motion — each governed by the same vector kinematics equations.

Course Competencies and Learning Objectives

A ★ indicates that this page contains content related to that LO.

  DRAFT: Learning objectives below are placeholders pending CME confirmation.

CC2.1 Analyze the components of motion and any applied forces

★ LO2.2.4 Apply kinematic equations independently to each coordinate direction in multi-dimensional motion

★ LO2.2.5 Analyze projectile motion by separating horizontal (constant velocity) and vertical (constant acceleration) components

★ LO2.2.6 Determine the centripetal acceleration of an object undergoing uniform circular motion

★ LO2.2.7 Apply the Galilean velocity addition formula to solve relative motion problems in one and two dimensions

Required Reading

Click the blue buttons to go to the OpenStax reading assignments. Complete all sections before watching the videos.

4.3 Projectile Motion 4.4 Uniform Circular Motion 4.5 Relative Motion in One and Two Dimensions

 

Optional Reading

Explore More

Want deeper practice or alternative explanations for projectile motion, circular motion, or relative velocity? These free resources provide additional worked examples and simulations.

Media

Videos are organized by topic. Work through the pre-lecture and mini-lecture tabs for each topic before attempting the corresponding sample problems.

Video 1: Pre-Lecture — Projectile Motion Overview

Pre-Lecture: Projectile Motion — The Two-Axis Framework (OpenStax 4.3)

This video introduces projectile motion as a special case of 2D kinematics where the only acceleration is gravitational (downward). The key insight is the complete independence of horizontal and vertical motion.

  • Assumptions: no air resistance; g = 9.80 m/s2 downward
  • Horizontal: ax = 0, so vx = v0x = constant
  • Vertical: ay = −g = −9.80 m/s2
  • The trajectory is a parabola — derived from eliminating t between x(t) and y(t)

Time: Video pending — to be recorded by instructor

Video 2: Mini-Lecture — Projectile Motion Problem Setup

Mini-Lecture: Setting Up and Solving Projectile Motion Problems

This worked-example video walks through the systematic approach to any projectile problem: decompose the initial velocity, apply the kinematic equations on each axis, and combine results to find range, max height, and time of flight.

  • Initial velocity components: v0x = v0 cos θ, v0y = v0 sin θ
  • Equations: x(t) = v0xt; y(t) = v0yt − ½gt2
  • Finding time of flight (when y = 0 for ground-level launch)
  • Range formula R = v02 sin(2θ) / g — and why 45° gives maximum range

Time: Video pending — to be recorded by instructor

Video 3: Sample Problem — Projectile Motion 1

Sample Problem: Projectile Launched at an Angle from Ground Level

A complete worked solution for a classic projectile problem: a ball is launched at an angle above the horizontal, and you are asked to find the maximum height, time in the air, horizontal range, and velocity at impact.

  • Decompose initial velocity into v0x and v0y
  • Find time to reach apex: vy = 0 at apex → tapex = v0y/g
  • Calculate maximum height: ymax = v0y2 / (2g)
  • Find range and impact velocity (magnitude and angle below horizontal)

Time: Video pending — to be recorded by instructor

Video 4: Sample Problem — Projectile Motion 2

Sample Problem: Projectile Launched Horizontally from a Height

A worked solution for a projectile launched horizontally from an elevated position — a common variant where v0y = 0. Covers time to reach the ground and horizontal distance traveled.

  • Initial conditions: v0x = v0, v0y = 0, launch height = h
  • Time to fall: h = ½gt2 → t = √(2h/g)
  • Horizontal range: R = v0x × t
  • Final velocity components and impact angle below horizontal

Time: Video pending — to be recorded by instructor

Video 5: Pre-Lecture — Uniform Circular Motion

Pre-Lecture: Uniform Circular Motion — Centripetal Acceleration (OpenStax 4.4)

This video shows that a particle moving in a circle at constant speed is nonetheless accelerating — because its direction is always changing. The acceleration vector points toward the center and is called centripetal acceleration.

  • Uniform circular motion: constant speed v, radius r, period T
  • Centripetal acceleration: ac = v2 / r, directed toward center
  • Period and frequency: T = 2πr / v, f = 1/T
  • Speed in terms of radius and period: v = 2πr / T

Time: Video pending — to be recorded by instructor

Video 6: Sample Problem — Circular Motion

Sample Problem: Centripetal Acceleration of an Object in Circular Motion

A complete worked solution for a uniform circular motion problem, finding the centripetal acceleration from given speed and radius, then finding the force required to maintain circular motion (preview of Newton's second law applied in 2D).

  • Given: radius r and speed v or period T — find ac
  • Convert between period and speed if necessary: v = 2πr / T
  • Apply ac = v2 / r; confirm units (m/s2)
  • Direction of ac: always toward the center of the circle

Time: Video pending — to be recorded by instructor

Video 7: Pre-Lecture — Relative Motion in 1D

Pre-Lecture: Relative Motion in One Dimension (OpenStax 4.5)

This video introduces reference frames and the Galilean velocity addition formula in one dimension. You will see how the observed velocity of an object depends on the motion of the observer's reference frame.

  • Reference frames S and S′ — one moving at velocity vS′S relative to the other
  • Galilean velocity addition (1D): vPA = vPB + vBA
  • Subscript notation: vPA = velocity of P as seen from frame A
  • Classic examples: passenger on a train, boat in a river (1D case)

Time: Video pending — to be recorded by instructor

Video 8: Pre-Lecture — Relative Motion in 2D

Pre-Lecture: Relative Motion in Two Dimensions (OpenStax 4.5 continued)

Extending relative motion into two dimensions using vector addition. The velocity of the passenger relative to the ground equals the velocity of the passenger relative to the vehicle plus the velocity of the vehicle relative to the ground.

  • Vector form: vPA = vPB + vBA
  • Each velocity is a 2D vector — add component by component
  • Classic 2D example: airplane flying in a crosswind
  • Boat crossing a river at an angle — finding the actual heading needed to go straight across

Time: Video pending — to be recorded by instructor

Video 9: Sample Problem — Relative Motion 1D

Sample Problem: Relative Velocity in One Dimension

A complete worked solution for a 1D relative motion problem: two vehicles moving in the same or opposite directions, finding the velocity of one as seen by an observer in the other.

  • Identify the two reference frames and the object being observed
  • Set up subscript notation correctly: vPA = vPB + vBA
  • Assign signs carefully — motion in positive or negative x-direction
  • Interpret the answer: what does the observer actually see?

Time: Video pending — to be recorded by instructor

Video 10: Sample Problem — Relative Motion 2D

Sample Problem: Relative Velocity in Two Dimensions (Crosswind / River Crossing)

A complete worked solution for a 2D relative motion problem — a classic airplane-in-a-crosswind or boat-crossing-a-river scenario — showing how to use vector addition to find both the actual velocity and the direction the vehicle must point to achieve a desired path.

  • Set up vPA = vPB + vBA in vector form
  • Resolve each velocity into x and y components
  • Solve for the unknown velocity or heading angle
  • Report magnitude and direction of the resultant velocity

Time: Video pending — to be recorded by instructor

Practice and Apply - Conceptual

Classify Each Statement

Sort each statement into the correct motion category: Projectile Motion, Uniform Circular Motion, or Relative Motion.

Sort Items

Concepts to Sort

  • Horizontal velocity is constant throughout the motion.
  • The acceleration is always directed toward the center of the path.
  • The observed velocity depends on the reference frame of the observer.
  • Speed is constant but velocity is always changing direction.
  • The trajectory is parabolic when only gravity acts.
  • vPA = vPB + vBA
  • ac = v2 / r
  • Vertical and horizontal components are analyzed independently.

Projectile Motion

    Uniform Circular Motion

      Relative Motion

        Key Concept Check

        Click each card to flip it and test your understanding across all three topic areas.

        In projectile motion, why is the horizontal velocity constant?
        Answer

        Because the only force acting is gravity, which points downward (negative y-direction). There is no horizontal component of force or acceleration: ax = 0. With zero acceleration, vx = v0x = constant throughout the entire flight.

         

        Why does an object in uniform circular motion have a centripetal acceleration even though its speed is constant?
        Answer

        Acceleration is the rate of change of the velocity vector, not just speed. In circular motion, the direction of v is continuously changing even though |v| is constant. The rate of change of direction produces a centripetal (center-pointing) acceleration of magnitude ac = v2/r.

         

        At what launch angle does a projectile achieve maximum range? Why?
        Answer

        45°. The range formula is R = v02 sin(2θ) / g. The function sin(2θ) is maximized when 2θ = 90°, so θ = 45°. This angle provides the optimal balance between time of flight (maximized by high launch angle) and horizontal speed (maximized by low launch angle).

         

        What does the subscript notation vPA mean in a relative motion problem?
        Answer

        vPA means "the velocity of object P as measured by observer in frame A." The first subscript is the object being observed; the second is the reference frame doing the observing. The Galilean addition rule is: vPA = vPB + vBA — the middle subscripts (B) cancel.

         

        How do you find the centripetal acceleration if you are given the period T and radius r instead of speed?
        Answer

        First, find speed from period and radius: v = 2πr / T. Then apply ac = v2 / r = (2πr/T)2 / r = 4π2r / T2. So the combined formula is ac = 4π2r / T2 — useful when the period is given instead of speed.

         

        A ball is dropped and another is thrown horizontally from the same height. Which hits the ground first?
        Answer

        They hit at exactly the same time. The vertical motion is identical for both — both start with v0y = 0 and fall under the same gravitational acceleration. The horizontal throw only adds a horizontal velocity, which has no effect on the time to fall. This is the classic independence of horizontal and vertical motion.

         

        Practice and Apply - Computational

        Important: Key Equations for This Lesson

        Projectile Motion (g = 9.80 m/s2, no air resistance):

        • Horizontal: x(t) = x0 + v0xt   |   vx = v0x = v0 cos θ
        • Vertical: y(t) = y0 + v0yt − ½gt2   |   vy = v0y − gt   |   v0y = v0 sin θ
        • Range (ground-level launch): R = v02 sin(2θ) / g

        Uniform Circular Motion:

        • Speed: v = 2πr / T   |   Centripetal acceleration: ac = v2 / r = 4π2r / T2
        • Direction of ac: always toward the center of the circle

        Relative Motion:

        • vPA = vPB + vBA    (vector equation — use components in 2D)

        Practice Problem 1 — Projectile Motion

        A soccer ball is kicked from the ground with an initial speed of 25.0 m/s at an angle of 38.0° above the horizontal. (a) Find the horizontal and vertical components of the initial velocity. (b) Find the maximum height reached by the ball. (c) Find the total time the ball is in the air. (d) Find the horizontal range.

        Show Solution

        Given: v0 = 25.0 m/s, θ = 38.0°, g = 9.80 m/s2

        Part (a) — Initial velocity components:

        v0x = v0 cos θ = (25.0)(cos 38.0°) = (25.0)(0.788) = 19.7 m/s

        v0y = v0 sin θ = (25.0)(sin 38.0°) = (25.0)(0.616) = 15.4 m/s

        Part (b) — Maximum height (vy = 0 at apex):

        vy2 = v0y2 − 2gymax → ymax = v0y2 / (2g)

        ymax = (15.4)2 / (2 × 9.80) = 237.16 / 19.60 = 12.1 m

        Part (c) — Total time in air (ground-level launch, yfinal = 0):

        ttotal = 2v0y / g = 2(15.4) / 9.80 = 3.14 s

        Part (d) — Horizontal range:

        R = v0x × ttotal = (19.7)(3.14) = 61.9 m

        Check using range formula: R = v02 sin(2θ)/g = (625)(sin 76°)/9.80 = (625)(0.970)/9.80 = 61.8 m ✓

        Practice Problem 2 — Uniform Circular Motion

        A satellite orbits Earth in a circular orbit of radius 7.20 × 106 m with a period of 6025 s. (a) Find the orbital speed of the satellite. (b) Find the centripetal acceleration of the satellite. (c) In which direction does the acceleration vector point?

        Show Solution

        Given: r = 7.20 × 106 m, T = 6025 s

        Part (a) — Orbital speed:

        v = 2πr / T = 2π(7.20 × 106) / 6025

        v = (4.524 × 107) / 6025 = 7.51 × 103 m/s = 7510 m/s

        Part (b) — Centripetal acceleration:

        ac = v2 / r = (7510)2 / (7.20 × 106)

        ac = (5.640 × 107) / (7.20 × 106) = 7.83 m/s2

        (Alternatively: ac = 4π2r/T2 = 4π2(7.20×106)/(6025)2 = 7.83 m/s2 ✓)

        Part (c) — Direction of acceleration:

        The centripetal acceleration vector always points toward the center of the orbit — in this case, toward Earth's center. This is the gravitational acceleration providing the centripetal force.

        Practice Problem 3 — Relative Motion in 2D

        An airplane heads due east with an airspeed of 240 km/h. A wind blows from south to north at 55.0 km/h. (a) Find the velocity of the airplane relative to the ground in unit-vector form. (b) Find the speed of the airplane relative to the ground. (c) Find the direction angle of the actual flight path measured from east.

        Show Solution

        Given: vplane/air = 240î km/h (east); vair/ground = 55.0ĵ km/h (north)

        Part (a) — Velocity of airplane relative to ground:

        vplane/ground = vplane/air + vair/ground

        vplane/ground = 240î + 55.0ĵ km/h

        Part (b) — Speed relative to ground:

        v = √(2402 + 55.02) = √(57600 + 3025) = √60625 = 246 km/h

        Part (c) — Direction angle north of east:

        θ = tan−1(vy / vx) = tan−1(55.0 / 240) = tan−1(0.229) = 12.9° north of east

        The wind pushes the plane northward, so the actual ground track deviates 12.9° north of due east even though the plane is pointing due east.

        Ready to Move On?

        Lesson Checklist

        Before moving on to M2L3, confirm you can do each of the following:

        • ☐ Decompose an initial velocity into horizontal and vertical components using v0x = v0 cos θ and v0y = v0 sin θ.
        • ☐ Solve for time of flight, maximum height, and horizontal range in a projectile motion problem.
        • ☐ Explain why a horizontally thrown ball and a dropped ball hit the ground at the same time.
        • ☐ Calculate centripetal acceleration using ac = v2/r or ac = 4π2r/T2.
        • ☐ Explain why an object in uniform circular motion is accelerating even though its speed is constant.
        • ☐ Apply vPA = vPB + vBA to find relative velocities in 1D and 2D problems.

        Projectile motion and circular motion will appear again when you apply Newton's second law in two dimensions — mastering these kinematics now will make Newton's Laws much more approachable!