Projectile & Two-Dimensional Motion · Grade 12 Science

Chapter: Motion & Kinematics

Projectile & Two-Dimensional Motion

Learning goals

Readings for this unit

Before this unit

Components, gravity, and quadratic time relationships.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Why does a ball keep moving forward after it leaves the hand?

Illustrative evidence case. A modeled ball follows a curved path while equal-time horizontal intervals stay equal.

A curved trajectory combines horizontal and vertical motionA curved trajectory combines horizontal and vertical motionHeight above launch y (m)051015010203040Horizontal position x (m)Green: vx; red: vy. Gravity remains downward at the apex.
Calculated trajectory x=12t, y=16t−4.9t²; arrow directions illustrate components, not additional forces.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Why does a ball keep moving forward after it leaves the hand?

Your first model

Draw velocity components separately from the one downward gravity force.

Build the language from the idea

projectile
An object moving through the air acted on only by gravity after its launch (air resistance neglected).
trajectory
The path a projectile follows, which is a parabola.
horizontal component
The part of a projectile's motion parallel to the ground; its velocity here stays constant.
vertical component
The part of a projectile's motion along the direction of gravity; it accelerates downward at g.
range
The horizontal distance a projectile travels before returning to its launch height.
independence of motion
The principle that a projectile's horizontal and vertical motions occur separately and do not affect each other's timing.

A projectile is an object whose modeled post-release motion is governed by gravity alone. Near Earth, approximately uniform gravity and negligible drag give a parabolic trajectory when horizontal velocity is nonzero. Real balls and water jets can experience drag, spin and breakup, so their paths need not be exact parabolas. Separate horizontal and vertical equations for the specified ideal model and then test how well they describe the observations.

For an ideal near-Earth projectile with negligible drag and approximately uniform gravity, horizontal acceleration is zero and vertical acceleration is downward with magnitude g. Component equations can be solved separately but share the same elapsed time and describe one motion. Real drag can couple their behavior. The component model is a specified approximation, not a universal claim that horizontal and vertical behavior can never interact.

A curved trajectory combines horizontal and vertical motionA curved trajectory combines horizontal and vertical motionHeight above launch y (m)051015010203040Horizontal position x (m)Green: vx; red: vy. Gravity remains downward at the apex.
Calculated trajectory x=12t, y=16t−4.9t²; arrow directions illustrate components, not additional forces.

Consider two identical model balls starting at the same height with the same zero initial vertical velocity. One is released vertically and one has an added horizontal velocity. Under uniform gravity with no drag and the same horizontal landing surface, their vertical position histories and landing times match. The horizontally launched ball lands farther away. Unequal heights, drag, or a changing terrain surface can invalidate this equal-time comparison.

This is why the strategy for every projectile problem is the same: split the motion into horizontal and vertical parts, and let time be the one quantity they share. Handle the horizontal part with constant-velocity motion, x = (horizontal velocity)(t). Handle the vertical part with the free-fall kinematic equations, using a = g. Consider a ball thrown horizontally at 15 m/s from a 19.6 m cliff. Vertically it falls 19.6 = ½(9.8)t², so t² = 4.0 and t = 2.0 s. Horizontally, in that same 2.0 s, it travels x = (15)(2.0) = 30 m before landing.

Use the same flight time in both perpendicular componentsUse the same flight time in both perpendicular componentsStepRelationshipResultFind landing timet=2vy₀/g3.27 sHorizontal travelx=vxt39.2 mLanding velocity(12,−16) m/sSpeed 20 m/sA shared clock connects the two component models.
Original constant-g, no-drag calculation; range and speed are model results, not measured throws.

When a projectile is launched at an angle instead of horizontally, you first split its launch velocity into components: a launch speed v₀ at angle θ above the horizontal gives a horizontal part v₀cosθ and a vertical part v₀sinθ. The horizontal part stays constant; the vertical part behaves like a ball thrown straight up, slowing to zero at the peak and speeding back up on the way down. On level ground the range works out to R = v₀²sin(2θ) ÷ g, an expression that is largest when θ = 45°, because sin(2θ) reaches its maximum of 1 there. That's why a 45° launch, in the absence of air resistance, throws something the farthest.

Solve vertical motion for the event the question actually asksSolve vertical motion for the event the question actually asksHeight above launch (m)0510150123Time since launch (s)Landing event: y=0 has t=0 and t=32/9.8 s.
Calculated vertical model; plotted event points follow the same launch components as the trajectory.

The ideal component model is useful because both directions share one clock while having separate acceleration equations. Drag can alter both velocity components and change the shape from a parabola; it does not merely tilt an otherwise unchanged parabola. The 45° maximum-range result additionally assumes a fixed launch speed, equal launch/landing heights, level landing surface, uniform gravity and negligible drag. Compare real evidence with those conditions before extending the result.

Model check: Horizontal motion does not cancel gravity; independence assumes negligible air resistance. Δx=vxt=6 m. Vertical fall is½gt²≈19.6 m withg=9.8 m/s².

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Resolve launch velocity into independent components

A projectile in this model moves after launch with gravity as its only force. Ignoring air resistance near Earth gives ax=0 and ay=−9.8 m/s² when up is positive. The plotted launch components are vx=12 m/s and vy₀=16 m/s. Horizontal velocity stays constant, while vertical velocity decreases, reaches zero and becomes negative. At the apex the object still moves horizontally and still accelerates downward; a curved path is not evidence of a forward force.

A curved trajectory combines horizontal and vertical motionA curved trajectory combines horizontal and vertical motionHeight above launch y (m)051015010203040Horizontal position x (m)Green: vx; red: vy. Gravity remains downward at the apex.
Calculated trajectory x=12t, y=16t−4.9t²; arrow directions illustrate components, not additional forces.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

At the apex, which velocity component is zero and which acceleration component is nonzero?

Visual reasoning — after your attempt

vy=0, vx=12 m/s and ay=−9.8 m/s². Total velocity and acceleration are not zero.

Choose horizontal and vertical axes and resolve the initial velocity using the stated launch angle. Under an ideal near-Earth model without drag, horizontal acceleration is zero and vertical acceleration is −g. Both components share the same elapsed time and together describe one trajectory.

Equal-time samples separate motion from gravityEqual-time samples separate motion from gravityHeight (m)024605101520Horizontal position (m)Red positions every 0.5 s; gold gravity arrows all point down.
Calculated x=10t,y=10t−4.9t² positions with constant downward acceleration and negligible drag. Equal horizontal spacing does not mean equal total speed; vertical velocity changes. Gold arrows indicate acceleration direction on a separate scale, not additional downward displacements.

Horizontal velocity can remain constant while vertical velocity changes. This is an approximation for a suitable object and range, not a universal feature of every flight.

Pause and explain

Are the horizontal and vertical motions separate trips?

Check your reasoning after answering

No. They are components of the same motion and occur simultaneously, so the time used in both calculations must match.

Part 2 · Solve vertical motion for the relevant event

An event condition identifies the target, such as reaching a height or touching the launch level again. Here y=16t−4.9t²=0 has roots t=0 and t≈3.27 s. The first is launch; the second is the later return. At the apex vy=16−9.8t=0 instead, giving t≈1.63 s. A landing below or above launch requires its own height condition and does not generally have the same symmetric flight time.

Solve vertical motion for the event the question actually asksSolve vertical motion for the event the question actually asksHeight above launch (m)0510150123Time since launch (s)Landing event: y=0 has t=0 and t=32/9.8 s.
Calculated vertical model; plotted event points follow the same launch components as the trajectory.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What is the maximum height above launch?

Visual reasoning — after your attempt

At vy=0, h=vy₀²/(2g)=256/19.6≈13.1 m.

Find flight time using the actual final height and launch height. The shortcut t=2 v₀y/g applies to return to the launch height under the ideal model. Landing above or below that height requires the vertical displacement equation and a physically appropriate root.

At maximum height vy=0, but total speed can remain nonzero because vx persists. Gravity still acts downward at the top.

Pause and explain

What fails when equal-height flight time is used for a cliff launch?

Check your reasoning after answering

The landing vertical displacement is not zero, so the equal-height symmetry and its time shortcut do not describe the event.

Part 3 · Use horizontal motion after finding time

The two components have separate equations but share the same elapsed time. Once the vertical equation gives the later launch-level landing at about 3.27 s, horizontal displacement is 12×3.27≈39.2 m. Recombining the landing components gives speed √(12²+16²)=20 m/s. This equality with launch speed follows equal height and negligible drag; it is not a universal rule for every landing elevation or real ball.

Use the same flight time in both perpendicular componentsUse the same flight time in both perpendicular componentsStepRelationshipResultFind landing timet=2vy₀/g3.27 sHorizontal travelx=vxt39.2 mLanding velocity(12,−16) m/sSpeed 20 m/sA shared clock connects the two component models.
Original constant-g, no-drag calculation; range and speed are model results, not measured throws.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Would doubling vx while retaining vy₀ change the modeled launch-level flight time?

Visual reasoning — after your attempt

No. The vertical equation is unchanged; range doubles while the same flight time applies.

Once the event time is known, Δx=v₀x t follows for constant horizontal velocity. For ideal equal-height launches, range depends on speed, angle, and g. A 45° maximum-range claim assumes the model’s conditions and does not automatically apply with drag or unequal elevations.

A larger horizontal component alone does not guarantee a longer range if the vertical component reduces flight time. Compare the complete two-component model.

Pause and explain

Why can a shallow launch fail to maximize range?

Check your reasoning after answering

It may have a large horizontal speed but too little time in the air. Range depends on the product of horizontal velocity and flight time.

Part 4 · Account for model limits and frame choice

A model limit identifies an excluded effect that could matter. Drag depends on relative air motion and can change both horizontal and vertical velocities; spin can add further aerodynamic effects. A moving observer also assigns a different horizontal velocity under ordinary frame transformations. The ideal parabola is a useful baseline for a stated interval, not an exact claim for every thrown ball. Compare observed positions with predictions and examine systematic deviations before adding complexity.

State drag, height and frame limits before extending a predictionState drag, height and frame limits before extending a predictionModel choiceWhat stays simpleWhat may changeNo drag, constant gvx constantParabolic trajectoryAir dragGravity still actsBoth components canchangeDifferent landingheightUse actual y targetFlight time/range differA reference-frame change also changes velocity labels.
Qualitative comparison of modeling assumptions; no invented drag trajectory is presented as measured evidence.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Does a noticeably shorter observed range prove gravity became stronger?

Visual reasoning — after your attempt

No. Drag, launch measurement, landing height or frame differences can also explain the mismatch; independent evidence is needed.

Drag, wind, changing elevation, and object shape can affect actual flight. The ideal model supplies a reference prediction, not certainty about a real experiment. Specify ground-frame velocities and ignore rotational or aerodynamic effects only when that simplification is appropriate.

Compare repeated measured landing positions with the ideal result and investigate a systematic shortfall. Do not adjust data to make them match the equation.

Pause and explain

What evidence could distinguish random scatter from a model shortfall?

Check your reasoning after answering

Repeated trials under documented conditions can show spread and consistent deviation. Measurements of drag-related conditions or another launch height can test competing explanations.

Apply the whole unit · Predict an ideal equal-height launch

Illustrative launch speed 20.0 m/s at 30° above horizontal; landing is at launch height. Use g=10.0 m/s², with no drag.

Your task

Find velocity components, time to the peak, maximum height, total flight time, and range. Explain what changes if the landing height is different.

What your response should show

  • Define the system, quantities, units, and relevant conditions before selecting a model.
  • Show intermediate calculations or a labeled evidence account, including signs or denominators.
  • Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
  • Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
  1. vx=20 cos 30°≈17.3 m/s; vy₀=20 sin 30°=10.0 m/s.

  2. Peak at t=1.00 s; height gain=vy₀²/(2 g)=5.00 m.

  3. Equal-height total time=2.00 s; range≈34.6 m.

These results share the same stated ideal model and g value. The speed at the peak is about 17.3 m/s horizontally, not zero. A different landing height changes the vertical event equation and therefore time and range; the simple doubled ascent time would no longer be generally valid.

Scientific references and source-model foundations

Parent / teacher teaching notes

Require the learner to identify the landing event before selecting a time formula. Use supplied data rather than throwing objects as an informal range test. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.

A curved trajectory combines horizontal and vertical motionA curved trajectory combines horizontal and vertical motionHeight above launch y (m)051015010203040Horizontal position x (m)Green: vx; red: vy. Gravity remains downward at the apex.
Calculated trajectory x=12t, y=16t−4.9t²; arrow directions illustrate components, not additional forces.

The key idea

Uniform gravity with negligible drag gives separate component equations sharing time. The 45° range maximum also assumes fixed launch speed and equal launch/landing heights.

Horizontal motion

x = (v₀·cosθ)·t

the horizontal velocity v₀·cosθ stays constant (no horizontal force, ignoring air resistance); for a horizontal launch θ = 0, so x = v₀·t.

Vertical motion

y = (v₀·sinθ)·t − ½·g·t²

the vertical part is free fall under g (taking up as positive); an object launched or dropped horizontally has v₀·sinθ = 0, so it simply falls ½·g·t².

Launch velocity components

horizontal = v₀·cosθ, vertical = v₀·sinθ

a launch speed v₀ at angle θ above the horizontal splits into independent horizontal and vertical parts.

Range on level ground

R = v₀²·sin(2θ) ÷ g

horizontal distance for a projectile landing at its launch height; maximized at θ = 45°, where sin(2θ) = 1.

Worked examples

Example 1

A ball is thrown horizontally at 15 m/s from the top of a 19.6 m cliff. Find (a) how long it is in the air and (b) how far from the base of the cliff it lands. (g = 9.8 m/s²)
  1. Vertical motion sets the time. Starting with zero vertical velocity, use height = ½gt²: 19.6 = ½(9.8)t² = 4.9t².
  2. t² = 19.6 ÷ 4.9 = 4.0, so t = 2.0 s.
  3. Horizontal motion is constant velocity: x = (15)(2.0) = 30 m.
Answer: (a) 2.0 s; (b) 30 m from the base

Example 2 · Worked application

A projectile leaves horizontally at 8 m/s from a 20 m height. Use g=10 m/s² and neglect drag. Find landing time and horizontal distance.
  1. Vertical motion starts with v_y=0: 20=½×10×t², so t=2 s.
  2. Horizontal velocity stays 8 m/s without horizontal force.
  3. Range=8×2=16 m.
Answer: 2 s and 16 m.
Check the reasoning

Solve vertical motion for time before using horizontal speed.

Example 3

A rock is dropped from rest and falls 44.1 m to the ground. How long does it take — and would a rock launched horizontally from that same height take longer? (g = 9.8 m/s²)
  1. Use 44.1 = ½(9.8)t² = 4.9t².
  2. t² = 44.1 ÷ 4.9 = 9.0, so t = 3.0 s.
  3. Because vertical motion is independent of horizontal motion, a horizontally launched rock falls in the same 3.0 s — its sideways speed doesn't change the timing.
Answer: 3.0 s; no — the horizontally launched rock takes the same 3.0 s

Example 4 · Find and fix the mistake

Context: A projectile leaves horizontally at 8 m/s from a 20 m height. Use g=10 m/s² and neglect drag. Find landing time and horizontal distance. A student claims: “A faster horizontal launch must keep the object airborne longer from the same height.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Solve vertical motion for time before using horizontal speed.
  2. Without drag and with the same initial vertical velocity, fall time is set by vertical motion, not horizontal speed.
Answer: Without drag and with the same initial vertical velocity, fall time is set by vertical motion, not horizontal speed.
Check the reasoning

Solve vertical motion for time before using horizontal speed.

Example 5 · Reason from the visual model

Use the same flight time in both perpendicular componentsUse the same flight time in both perpendicular componentsStepRelationshipResultFind landing timet=2vy₀/g3.27 sHorizontal travelx=vxt39.2 mLanding velocity(12,−16) m/sSpeed 20 m/sA shared clock connects the two component models.
Original constant-g, no-drag calculation; range and speed are model results, not measured throws.
Would doubling vx while retaining vy₀ change the modeled launch-level flight time?
  1. The two components have separate equations but share the same elapsed time. Once the vertical equation gives the later launch-level landing at about 3.27 s, horizontal displacement is 12×3.27≈39.2 m. Recombining the landing components gives speed √(12²+16²)=20 m/s. This equality with launch speed follows equal height and negligible drag; it is not a universal rule for every landing elevation or real ball.
  2. No. The vertical equation is unchanged; range doubles while the same flight time applies.
Answer: No. The vertical equation is unchanged; range doubles while the same flight time applies.
Check the reasoning

Original constant-g, no-drag calculation; range and speed are model results, not measured throws.

Avoid these mistakes

Check your understanding

A horizontal launch lasts 2 s at vx=3 m/s. Find horizontal range.

Discuss the answer

Δx=vxt=6 m. Vertical fall is½gt²≈19.6 m withg=9.8 m/s².

Return to the phenomenon

Return to the initial case. Why does a ball keep moving forward after it leaves the hand? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Inertia preserves horizontal velocity without horizontal net force; gravity changes the vertical component. At the top only vertical velocity vanishes.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Why does a ball keep moving forward after it leaves the hand? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the physical/environmental mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep object/material identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Inertia preserves horizontal velocity without horizontal net force; gravity changes the vertical component. At the top only vertical velocity vanishes.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

A ball is launched horizontally at 8.0 m/s from a 19.6 m ledge. Ignore drag and use g=9.8 m/s². Find flight time, horizontal range and landing velocity components.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

0=19.6−4.9t² gives t=2.00 s. Range=8×2=16.0 m. Landing velocity=(8.0,−19.6) m/s in right/up-positive axes; speed≈21.2 m/s.

Words to know

projectile
An object moving through the air acted on only by gravity after its launch (air resistance neglected).
trajectory
The path a projectile follows, which is a parabola.
horizontal component
The part of a projectile's motion parallel to the ground; its velocity here stays constant.
vertical component
The part of a projectile's motion along the direction of gravity; it accelerates downward at g.
range
The horizontal distance a projectile travels before returning to its launch height.
independence of motion
The principle that a projectile's horizontal and vertical motions occur separately and do not affect each other's timing.

Galileo's great insight: motion splits in two

Before Galileo, thinkers imagined a cannonball flew in a straight line until it 'ran out of push' and dropped. Galileo's revolutionary claim was that the horizontal and vertical motions are entirely independent: the ball moves forward at a steady rate and falls exactly as a dropped ball would, at the same time, and combining the two produces the parabola we actually see. This decomposition — treating one messy two-dimensional motion as two simple one-dimensional motions sharing a clock — is one of the most powerful problem-solving moves in all of physics, and it's the reason projectile problems are tractable at all.

Why 45° gives maximum range — and why real sports don't use it

On level ground and ignoring air resistance, the range formula R = v₀²sin(2θ)/g is largest when sin(2θ) = 1, which happens at θ = 45°. That's the textbook answer, and complementary angles (say 30° and 60°) give equal ranges because sin(2θ) is the same for both. But the real world adds wrinkles: air resistance flattens the ideal parabola and lowers the best angle, and in sports the launch point isn't at ground level. A shot-putter releases the shot from above the ground, which drops the optimal angle to roughly 42°, and a long-jumper is limited by how much vertical speed they can generate, so they take off near 20°. The ideal model is the starting point, not the final word.

The monkey and the hunter

One of physics' most beloved demonstrations imagines a hunter aiming a dart directly at a monkey hanging from a branch. The catch: the monkey lets go at the exact instant the dart is fired. Novices assume the hunter should aim high to compensate for the dart's fall — but the correct move is to aim straight at the monkey, because the dart and the monkey fall at the same rate once airborne. In the time the dart flies to the monkey's original position, gravity pulls it below the aim line by ½gt² — exactly the distance the monkey has dropped. They always meet, at any dart speed. It's the independence of horizontal and vertical motion made vivid, and it works every time in the lecture hall.

Beyond the ideal parabola: drag, spin, and real trajectories

The clean parabola assumes gravity is the only force, but real projectiles push through air. Air resistance (drag) grows with speed and steadily saps horizontal velocity, so a real trajectory is lopsided — rising at a shallower-than-parabola arc and falling more steeply, with a shorter range than the ideal formula predicts. Spin adds even more: a curveball or a hooked golf shot bends sideways because spinning surfaces drag air around unevenly, creating a sideways force called the Magnus effect. Ballistics experts, ski-jump designers, and sports engineers start from the ideal projectile model in this lesson, then layer drag and spin on top to match what actually happens in the air.

In the real world

Whether it's a quarterback leading a receiver, an artillery crew setting an angle, or an animator drawing a jump, projectile analysis predicts where a launched object will land — and it's why coaches teach a launch angle near 45° for maximum distance.

Exit check

A horizontal launch lasts 2 s at vx=3 m/s. Find horizontal range.

Check your explanation after answering

Δx=vxt=6 m. Vertical fall is½gt²≈19.6 m withg=9.8 m/s².

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

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