Two Dimensional Motion

Physics

Maple Hill High School 2025-26

Free Fall Motion

Free Fall

An object in free fall experiences only the force of gravity.

  • This means that it accelerates at , which is referred to as "acceleration due to gravity" or "gravitational field strength"
  • Units N/kg and m/s/s are the same
  • We ignore air resistance unless a problem tells us not to

How Big Is ?

Near the surface of the Earth,

In AP Physics we will use

How Big Is ?

Near the surface of the Earth,

In AP Physics we will use

is a magnitude, so it is always positive. The sign of the acceleration comes from the direction you chose as positive — not from itself.

Pick a Positive Direction First

If up is positive

  • Rising is
  • Falling is
  • Landing below the start is

If down is positive

  • Falling is
  • Rising is
  • Landing below the start is

Both choices give the same physics. Pick one, write it on your paper, and keep it for the whole problem. Most free fall mistakes are sign errors from switching halfway through.

Free Fall Facts

If an object is dropped,

Free Fall Facts

If an object is projected upwards it will slow down as it rises. The y-velocity will be zero at its highest point, or peak.

You can treat at the peak

Free Fall Facts

Up and down are mirror images

An object projected upward returns to its initial height with a velocity equal in magnitude and opposite in direction.

A ball thrown up at m/s passes its launch height again at m/s.

Up is positive  •  m/s/s

+30 0 −30 3.0 s 6.0 s velocity (m/s) time (s) area = +45 m rising area = −45 m falling slope = a = −10 m/s² v = 0 at the peak

Two Graphs, One Throw

45 0 3.0 6.0 position (m) time (s) position vs. time
+30 0 −30 3.0 6.0 velocity (m/s) time (s) velocity vs. time

Discuss — The position graph is curved but the velocity graph is straight. What does that tell you about the acceleration? Where on the position graph is the velocity zero?

The Cross Diagram

Five variables describe one stage of constant-acceleration motion. The cross puts them where you can see how they relate.

  • Top row is space, bottom row is time
  • Left column is the start, right column is the end
  • Middle column is what changes

After Sinkovits, Phys. Teach. 60, 744 (2022)

Δy v0 a v Δt space time start end the change

Five Boxes, Five Equations

Δy v0 a v Δt left Δy = v0Δt + ½aΔt² Δy v0 a v Δt right Δy = vΔt − ½aΔt² Δy v0 a v Δt top v² − v0² = 2aΔy Δy v0 a v Δt bottom v − v0 = aΔt Δy v0 a v Δt outside (no a) Δy = ½(v0 + v)Δt

How to use it. Underline every known. Star the unknown you want. Then find the box that holds all your knowns plus exactly one unknown — that box is the equation to use.

Problem-Solving Process

MOVE 1 — Picture & Direction. Sketch it. Choose which way is positive and write it down — then set to match.

MOVE 2 — Graphs. Sketch the shape of the – and – graphs.

MOVE 3 — Organize. Fill in the cross. Underline every known — with its sign. Star the one you are solving for.

MOVE 4 — Plan & Solve. Draw the box that holds all your knowns plus exactly one unknown. That box is your equation. Substitute, then solve.

MOVE 5 — Check. Sign, size, units.

Worked Examples

Free Fall with

Our toolkit:

A ball is dropped from rest from . Up is positive. Find:

  • time to impact
  • impact velocity (with sign)











A rock is thrown straight up at from ground level.

  • Time to the top. Hint: at peak.
  • Height at the top.











A ball is thrown downward at from a height of .

  • How long does it take to reach the ground?











Drop from . Down is positive. What is the impact velocity?

Template:











Compare: does a head start change the fall?

Case A (drop): from

Case B (throw down): from









Error Analysis — Sign Convention

A student writes: up is positive, , for a ball thrown upward at — and concludes the ball speeds up on the way up.

Diagnose: with up positive, . Velocity and acceleration point in opposite directions, so the ball must slow down as it rises.

Correct the peak time:

Quick Checks Before You Commit

  • Which equation matches your knowns and unknowns?
  • Are all signs consistent with the positive direction you chose?
  • Does the magnitude make physical sense? Is the unit right?
  • If the object went up and came back, is your time about double the time to the peak?

Heads Up: CalcPad Uses

Physics Classroom auto-grades with . Use 9.8 on the CalcPad sets so your answers are accepted. Use 10 on AP-style problems, quizzes and the exam.

This is the one place in the unit where the two numbers matter. Everywhere else, 10 is close enough and much faster without a calculator.

Vectors

↖️ ↗️ ↘️ ⬆️ ⬇️

Vector Quantities:

A vector is a quantity with both magnitude (size) and direction.

Examples:

  • The child was displaced 9 meters North.
  • The car has a velocity of 10 meters per second East
  • , , , , , etc

Scalar Quantities

A scalar is a quantity with just magnitude.

Examples

  • The child traveled a distance of 12 meters
  • The car is moving 20 miles per hour
  • The frog has mass of 0.5 kg.
  • , , , etc

Representing Vectors

10 20 30 40 10 20 x y |a| = 51.5

Representing Vectors

  • a scale is clearly listed
  • a vector arrow (with arrowhead) is drawn in a specified direction. The vector arrow has a head and a tail.
  • the magnitude and direction of the vector is clearly labeled.
    • The magnitude is 20 m and the direction is (30 degrees West of North).
30° d = 20 m north east SCALE: 1 cm = 4 m 20 m, 30° west of north

Vector Direction Conventions

  1. expressed as an angle of rotation about its "tail"
    • 40 degrees North of West (meaning a vector pointing West has been rotated 40 degrees towards the northerly direction)
    • 65 degrees East of South (meaning a vector pointing South has been rotated 65 degrees towards the easterly direction).
  2. expressed as a counterclockwise angle of rotation about its "tail"
    • 30 degrees is a vector that has been rotated 30 degrees in a counterclockwise direction relative to due east
    • 160 degrees is a vector that has been rotated 160 degrees in a counterclockwise direction relative to due east

Vector Direction Conventions

40° 40° counterclockwise from east 240° 240° counterclockwise from east 0° is due east; angles open counterclockwise
Direction = 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360° counterclockwise from east

The Magnitude of a Vector

SCALE: 1 cm = 5 miles 60° d = 15 mi 60° d = 25 mi 60° d = 18 mi same direction — the arrow's length carries the magnitude
  • Magnitude of a vector is scaled in the vector diagram by the length of the arrow
  • This can be scaled (i.e. 1 cm = 5 miles)

Adding Vectors

  • When vectors are in the same, or opposite directions we can add them with regular algebra
  • Line them up on one axis, keep the signs, and the arithmetic is the vector sum
5 + 5 = 10 5 + −5 = 0 5 + 10 = 15 5 + −10 = −5 5 + −15 = −10 10 + −5 = 5

Try It...

Mr. Porter and his wife walk from Nighthawks to Troy Savings Bank Music Hall. They walk 6 blocks East and then 2 Blocks South. [1 Block = 100 Meters]

  1. Determine the distance that they traveled.
  2. Determine their displacement.

Adding Angled Vectors

10 N + 10 N = ? ? ? 10 N + 10 N = ? ? ? 10 N + 5 N = ? ? ? not collinear — algebra alone will not do it

Adding Vectors

Vectors are added "tip to tail", that is redraw the vectors so that the tip of one vector is attached to the tail of the second vector.

The resultant vector S is equal to the addition of vectors a and b


Right Angled Vectors

Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement.

11 km, N + 11 km, E = 11 km, N 11 km, E R the legs meet at a right angle, so R closes the triangle

Determining the Angle

a b c θ a adjacent · b opposite · c hypotenuse

11 km, N 11 km, E R = 15.6 km θ θ is measured from the north leg

The Calculated Angle isn't always the answer...

10 km, West  +  5 km, South

10 km, W 5 km, S R 26.6° 180° 26.6° R direction of R = 180° + 26.6° = 206.6° 26.6° is measured off the west leg

Using Scaled Vectors

SCALE 3 cm 5 cm 8 cm a longer arrow means a larger magnitude Home Base every leg drawn to the same scale, tip to tail

Scaled Vectors

45° 300° 210° + + angles measured counterclockwise from east  ·  arrows drawn to scale

Scaled Vectors

45° 300° 210° + + angles counterclockwise from east
20 m 45° 25 m 300° 15 m 210° 20.3 m at 312° resultant: tail of the first to the tip of the last

Resultant: the vector sum of two or more vectors. It is the result of adding two or more vectors

  • Resultant Displacement or Resultant Force or Resultant Velocity

Order Does Not Matter

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Stop 🛑

Representing Vector Components

10 20 30 40 10 20 x y 45.0 25.0 |a| = 51.5

Vector Components

  • Parts of a two-dimensional vector
  • The component of a vector is the influence of that vector in a given direction.
    • i.e. How far East of a North East displacement did you walk?
  • We look at the perpendicular components
    • How much of the vector is in the x-direction
    • How much of the vector is in the y-direction
  • Vector is made up of components and

Vector Components Math

  • Notice this is a right triangle
    • Depending on the angle...
10 20 30 40 10 20 x y 45.0 25.0 |a| = 51.5

What are the components of Vector ?

10 20 30 10 x y 23.2° |a| = 38.1

What are the components of Vector and ?

10 20 10 x y 36.9° 108.4° |a| = 25.0 |b| = 15.8

Use Trig to Find and

10 20 30 10 x y 25.7° |a| = 30.0

Adding Vectors

Using Components

You can sum the components of the two vectors to find the components of the resultant vector

If ...

Adding Perpendicular Vectors

Mac and Tosh are doing the Vector Walk Lab. Starting at the door of their physics classroom, they walk 2.0 meters, south. They make a right hand turn and walk 16.0 meters, west. They turn right again and walk 24.0 meters, north. They then turn left and walk 36.0 meters, west. What is the magnitude of their overall displacement?

A graphical representation of the given problem will help visualize what is happening. The diagram below depicts such a representation.

2.0 m, S 16.0 m, W 24.0 m, N 36.0 m, W + + + the four legs, each drawn to the same scale

Step 1: Tip-to-tail

2.0 m, S 16.0 m, W 24.0 m, N 36.0 m, W R walk the legs tip to tail — R joins start to finish

Step 2: Sum and directions

2.0 m, S 24.0 m, N 16.0 m, W 36.0 m, W R 52.0 m, W total 22.0 m, N total group the north–south legs, then the east–west legs

Practice 🎯

Adding 2 Perpendicular Vectors

Adding 3 or More Perpendicular Vectors

Vector Treasure Hunt 🪙

  1. Create a Vector Map (draw this on the map of the school) -> start at one of the entrances of the school.
  2. Record all vectors in component form and Magnitude-direction from (use the convention degrees clockwise from North...this is the setting your iPhone compass uses)
  3. Exchange vector directions with another group and draw their vector map on your grid in another color or symbol (i.e. dotted line)
  4. Calculate the displacement of the the vector map
  5. Go walk the map and record the landmarks at the end of each vector.

Vector Descriptions

  • Component Form:
  • Magnitude-direction form:
    • units at degrees
10 20 10 x y 20.0 10.0 26.6° |a| = 22.4

Horizontal Projectiles

Two toy trucks roll off the ends of identical tables. The speeds and masses of the trucks are given.

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Will Truck A be in the air for (i) a longer time, (ii) a shorter time, or (iii) the same time as Truck B before it reaches the floor?

Explain your reasoning.

Two toy trucks roll off the ends of identical tables. The speeds and masses of the trucks are given.

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Will Truck A be in the air for (i) a longer time, (ii) a shorter time, or (iii) the same time as Truck B before it reaches the floor?

Explain your reasoning.

Launcher Demo

What are the three possible outcomes?

  1. Right ball () hits first
  2. Left ball () hits first
  3. Balls hit at the same time

What does each result imply about the horizontal velocity's effect on flight time?

  1. Right ball () hits first
  2. Left ball () hits first
  3. Balls hit at the same time

What the Demo Showed

The two motions are independent

Both balls are released at the same instant. Every dashed line joins them at the same moment in time.

They are always at the same height — the horizontal velocity does nothing to the fall.

Positions drawn at equal time steps

released together dropped launched

Horizontal Projectiles

  • A projectile is an object that only experiences a gravitational force
  • For horizontal projectiles the
  • We can separate motion in the and planes and solve separately

Path of a Projectile

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Path of a Projectile


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Horizontal Motion Vertical Motion
Forces
(Present? - Yes or No)
(If present, what dir'n?)		 No Yes
The force of gravity acts downward
Acceleration
(Present? - Yes or No)
(If present, what dir'n?)		 No Yes
is downward at 10 m/s/s
Velocity
(Constant or Changing?)		 Constant Changing
(by 10 m/s each second)

Horizontal and Vertical Velocity

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Horizontal and Vertical Displacement

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The Two Graphs That Matter

20 0 area = Δx = 60 m v (m/s) time (s) horizontal slope = 0, so no acceleration
30 0 area = Δy = 45 m v (m/s) time (s) vertical slope = +10 m/s²

Same clock, two independent stories. The horizontal graph is flat because nothing pushes sideways. The vertical graph is identical to a dropped object's.

Down is here, so climbs from zero and every number stays positive. The area under each graph is the displacement in that direction.

Horizontally Launched Projectiles

  • Projectile launched with an initial horizontal velocity from an elevated position.
  • Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile.

Horizontal Projectiles

X-Plane Motion

  • , so is constant

Y-Plane Motion

  • constant acceleration

The Double Cross Diagram

Two motions, one clock. Stack a horizontal cross on a vertical cross and let them share .

  • The cross has , so it collapses to
  • The cross is ordinary free fall
  • is the only quantity they share

That shared is the whole trick. Find the time in one cross, then carry it into the other.

The launcher demo is why that row is shared — the clock belongs to the fall, not to the launch.

Δx v0x ax vx Δt v0y ay vy Δy horizontal vertical shared by both

Three Boxes, Three Equations

Same rule as free fall: find the box that holds all your knowns plus exactly one unknown. With the horizontal cross collapses to a single box, so there are only three left.

Δx v0x ax vx Δt v0y ay vy Δy horizontal box Δx = vxΔt Δx v0x ax vx Δt v0y ay vy Δy vertical box Δy = v0yΔt + ½ayΔt² Δx v0x ax vx Δt v0y ay vy Δy vertical, without Δy vy = v0y + ayΔt

Problem-solving Process

MOVE 1 — Picture & Direction. Sketch it. Choose positive directions for and and write them down.

MOVE 2 — Graphs. Sketch – and –. One is flat; one is a slope.

MOVE 3 — Organize. Fill in the double cross. Underline every known. Star what you want.

MOVE 4 — Plan & Solve. Two beats: find in whichever cross has enough knowns, then cross over and use it in the other.

MOVE 5 — Check. Sign, size, units.

Δx v0x ax vx Δt v0y ay vy Δy horizontal vertical

Example

A ball rolls off a table 1.25 m high at 3.0 m/s.

How long is it in the air, and how far from the table's edge does it land?

Δx v0x ax vx v0y ay vy Δy Δt

Example — Two Beats

Δx = ? 3.0 0 3.0 0 +10 — +1.25 ★Δt beat one — find Δt 1.25 = ½(10)Δt² Δt = √0.25 = 0.50 s
★Δx 3.0 0 3.0 0 +10 — +1.25 0.50 beat two — cross over Δx = vxΔt = (3.0)(0.50) Δx = 1.5 m

The shared did the work. The 3.0 m/s never entered beat one — the ball falls exactly as if it had been dropped.

Practice 1 — off the edge

A ball rolls off a counter 0.80 m high at 2.5 m/s.

Find the time in the air and the horizontal distance it lands from the counter.

Practice 1 — answer

Beat one — vertical box

Beat two — cross over

The 2.5 m/s never touched beat one — the ball fell exactly as if dropped.

Practice 2 — back to the launch speed

A marble rolls off a 20 m cliff and lands 24 m from the base.

Find its launch speed.

Practice 2 — answer

Beat one — vertical box

Beat two — cross over

Both lengths were given, so the vertical cross still had enough on its own to give .

Practice 3 — starting in the other cross

A rock is thrown horizontally at 6.0 m/s and hits the water 18 m from the base of the cliff.

How high is the cliff?

Which cross has enough knowns to give you this time?

Practice 3 — answer

Beat one — horizontal box

Beat two — cross over

This one started in the other cross — the horizontal one was the complete one. Count what you were given in each cross before you write anything.

Putting the Components Back Together

You just found and at impact. They are perpendicular, so you already know how to combine them — this is the vector work from the start of the unit.

Not by adding the numbers. .

vx = 20 m/s vy = 30 m/s v = 36 m/s θ = 56° √(20² + 30²) = √1300 = 36 m/s, 56° below the horizontal

Note what did not change. is still the 20 m/s it was launched with — gravity never touched it. All of the extra speed at impact came from , built up over the fall.

Practice 4 — the whole thing

A rock is kicked horizontally off a 45 m cliff and lands 60 m from the base.

  1. Find the launch speed.
  2. Find the velocity — speed and direction — with which it hits the ground.

Practice 4 — answer

Beat one — vertical box

Beat two — the two components

Put them back together

below the horizontal

is still the 20 m/s it launched with — gravity never touched it.

Practice 5 — find the error

The problem. A ball rolls off a 1.25 m table at 3.0 m/s. How far from the edge does it land?

A student's work. "The ball starts at 3.0 m/s, so m/s. Also and ."

The student made one mistake. Name it, and say which cell of the double cross it lands in.

Practice 5 — answer

The mistake. They put the launch speed into . The launch is horizontal, so — the 3.0 m/s belongs in the top cross as , and nowhere in the bottom one.

Done correctly

Why their time came out too short. They gave the ball a downward head start it never had. Starting the fall at 3.0 m/s instead of 0 covers the 1.25 m sooner — so the time is too small, and the range with it.

🔑 Key Ideas:

A projectile is an object with the following characteristics:

  • The only force acting on it is the gravitational force; it is a free-falling object
  • The acceleration is directed downwards and has a magnitude of (we use on CalcPad)
  • Once projected, it continues its horizontal motion without any need of a force
  • As it travels through the air, its horizontal velocity remains constant
  • As it rises, its vertical velocity () decreases; as it falls, its increases

Marble Launch Lab

If you change only the launch height, how does the range depend on it?

Marble Launch Lab — the job

Purpose. Derive how range depends on launch height, test it on a hand-drawn linearized graph, and use the slope to measure .

Derive first: eliminate between the two crosses.

The data

  • 8×10 rule — at least 8 launch heights, the largest at least 10× the smallest (~10 cm to 1.0 m)
  • measured floor → launch point; every from the plumb mark directly below it
  • from the photogate: , marble diameter by calipers, repeated

The analysis

  • Plot against by hand — slope off the line, not off two data points
  • , then

Three things that will wreck your data

1. The ramp travels with the platform. Same release point, same tilt, every trial. Change the angle and changes — and the whole derivation assumes it did not.

2. Measure to the launch point, not to the table.

3. Set the gate beam across the marble's widest part. Clip it high or low and the beam cuts a short chord, so is short and reads low.

Safety: stray marbles are a slip hazard — collect them. Weight the ring stand base. Step stool, never a chair.

Angled Projectiles

Splitting the Launch Velocity

A projectile launched at an angle has one velocity, but it is easier to work with its two perpendicular components.

Do this first, always. Every angled projectile problem starts by splitting into and .

θ v 0 v 0x v 0y

Components

  • The X-axis velocity component () is constant and will not accelerate, so it stays the same for the entire flight
  • The initial Y-axis velocity component changes, because it is accelerated by gravity at 10 m/s² down. This is why we write () and not just — it will change throughout the problem.

Finding x & y components of initial velocity

Velocity Through the Flight

peak: vy = 0 vy shrinks vy grows, now downward vx never changes positions drawn at equal time steps

Cannonballs of different masses are shot from cannons at various angles above the horizontal. The velocity of each cannonball as it leaves the cannon is given, along with the horizontal component of that velocity, which is the same.

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Rank the horizontal distance traveled by the cannonballs.

Cannonballs with different masses are shot from cannons at various angles above the horizontal. The velocity of each cannonball as it leaves the cannon is given, along with the same vertical component of that velocity.

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Rank the time the cannonballs are in the air.

Projectile Fact Reminders

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Carried Over From Free Fall

  • Up positive  •  down positive
  • At the peak, — but is still downward
  • Launch and landing speeds match at the same height: up at means down at

Nothing new vertically. An angled projectile's -motion is the free fall you already solved — the only addition is a constant running alongside it.

Problem-solving Process

  1. Read & sketch problem
  2. Calculate x & y components of initial velocity
  3. Identify given information & fill out double cross diagram
  4. Identify the quantity you are solving for
  5. Using either horizontal or vertical information, find the flight time of the projectile
  6. With the time find the quantity you are solving for.

Where should 🎯 go?

Relative Motion

Relative Velocity and Vector Addition

  • Goal: Understand how velocities add as vectors and why motion is relative to the observer.

📝 Complete AP Workbook 1.E

1D Relative Motion: Vector Sums

plane velocity wind velocity resultant 100 km/hr + = 25 km/hr 125 km/hr tailwind — the two add
plane velocity wind velocity resultant 100 km/hr + = 25 km/hr 75 km/hr headwind — they subtract

Pivot: One-Dimensional Relative Motion Using a Drone

Boat Crossing a River: Resultant Velocity

When a boat heads straight across, the river current carries it downstream. The boat’s velocity relative to water adds vectorially to the current’s velocity relative to ground.

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plane velocity wind velocity resultant 100 km/hr + = 25 km/hr R R = 103.1 km/hr crosswind — perpendicular, so use Pythagoras

Riverboat Problems

  • River current is 4 m/s North
  • Boat velocity is 4 m/s East
  • What does a drone see?
  • How can we predict this resultant velocity?
boat velocity river current path over the ground = boat + current heading straight across, the boat still lands downstream

Example Riverboat

A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North.

  1. What is the resultant velocity of the motorboat?
  2. If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
  3. What distance downstream does the boat reach the opposite shore?
distance A — boat speed distance B river speed distance C — resultant start finish the same trip, read three ways: across, downstream, and the resultant

Riverboat Practice

Pivot: 2D Motion with a Drone

Reveal: - $v_0 = 0$, $a = -10$, $\Delta y = -20$ - $-20 = \tfrac12(-10)t^2 \Rightarrow t^2 = 4 \Rightarrow t = 2.0\ \mathrm{s}$ - $v = at = (-10)(2.0) = -20\ \mathrm{m/s}$ (20 m/s downward)

Reveal: - $t_{\text{top}} = \dfrac{0 - v_0}{-g} = \dfrac{12.0}{10} = 1.2\ \mathrm{s}$ - $y_{\max} = v_0 t + \tfrac12 a t^2 = 12(1.2) - 5(1.2)^2 = 7.2\ \mathrm{m}$ - Check with $y_{\max} = \dfrac{v_0^2}{2g} = \dfrac{144}{20} = 7.2\ \mathrm{m}$

Reveal: - $v_0 = 0$, $a = +10$, $\Delta y = +45.0$ - $v^2 = 2(10)(45.0) = 900 \Rightarrow v = +30\ \mathrm{m/s}$ (downward) - Notice: no time needed. This equation skips $t$ entirely.

Down is positive, $a = +10$, $\Delta y = +10.0$ for both. - A: $10 = \tfrac12 (10) t^2 \Rightarrow t^2 = 2 \Rightarrow t \approx 1.41\ \mathrm{s}$ - B: $10 = 5t + 5t^2 \Rightarrow t^2 + t - 2 = 0 \Rightarrow (t+2)(t-1) = 0 \Rightarrow t = 1.0\ \mathrm{s}$ Prompt: What stayed the same? What changed? (Same $a$, same $\Delta y$ — only $v_0$ differs.)