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SCIENCE · INVESTIGATE AND EXPLAIN

Translate a projectile between inertial observers

A projectile trajectory is a relationship between positions in a chosen frame. An observer moving horizontally at constant velocity u measures x′=x−ut if both origins coincide at launch. The vertical coordinate is unchanged because the observer has no vertical motion. Time is shared in this nonrelativistic model.

Learning outcome: Fix ground horizontal velocity at 5 m/s and compare observer velocities −3,3 and 5 m/s. Then choose a ground horizontal velocity below the observer velocity to make relative motion negative. Check that apex height, return time and vertical landing velocity remain unchanged.

Starting recap

Use signed quantities, algebraic substitution, graph slopes and areas. State the system, coordinate axes and units before calculating.

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Try

Let ground horizontal velocity be 5 m/s, upward velocity 10 m/s and g=10 m/s². Flight time is 2 s. A 3 m/s observer sees horizontal velocity 2 m/s and landing displacement 4 m. The ground range is 10 m. An observer moving at 5 m/s sees zero horizontal displacement throughout.

Open the experiment

Predict

Can the projectile move backward in one observer’s coordinates while moving forward relative to the ground?

Experiment

Fix ground horizontal velocity at 5 m/s and compare observer velocities −3,3 and 5 m/s. Then choose a ground horizontal velocity below the observer velocity to make relative motion negative. Check that apex height, return time and vertical landing velocity remain unchanged.

Describe a projectile from a moving observer

Worked teaching model · synthetic values · Point projectile has initial vertical velocity 10 m/s and g=10 m/s², returning to launch height after 2 s. Observer begins at launch point, translates horizontally at constant speed, and shares clock and axis orientation. No relativity or drag.

Reset keeps your prediction, captures and practice history.

01050horizontal position (m)height (m)

Axes rescale with settings; compare numerical values, not screen angles or apparent lengths. Solid/points show the selected model relationship or observations; the dashed comparison is explained below.

Calculated evidence ledger — synthetic teaching model
QuantityValueUnit
Return time2s
Apex time1s
Maximum height5m
Ground horizontal displacement10m
Observer horizontal displacement4m
Horizontal velocity in displayed frame2m/s
Vertical velocity at return-10m/s
Accessible plot coordinates
Primary plotted points
horizontal position (m)height (m)
00
0.20.95
0.41.8
0.62.55
0.83.2
13.75
1.24.2
1.44.55
1.64.8
1.84.95
25
2.24.95
2.44.8
2.64.55
2.84.2
33.75
3.23.2
3.42.55
3.61.8
3.80.95
40
Dashed comparison coordinates
horizontal position (m)height (m)
00
0.50.95
11.8
1.52.55
23.2
2.53.75
34.2
3.54.55
44.8
4.54.95
55
5.54.95
64.8
6.54.55
74.2
7.53.75
83.2
8.52.55
91.8
9.50.95
100

Changing the observer can reverse horizontal motion without changing vertical motion or acceleration.

Solve y=t(vy−gt/2)=0: the nonzero return time is 2vy/g. The projectile has zero vertical velocity at the apex but retains its horizontal velocity. Equal launch and landing height is essential for this time formula.

Changing the observer can reverse horizontal motion without changing vertical motion or acceleration.

Solve y=t(vy−gt/2)=0: the nonzero return time is 2vy/g. The projectile has zero vertical velocity at the apex but retains its horizontal velocity. Equal launch and landing height is essential for this time formula.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

1. Orient: Translate a projectile between inertial observers

A projectile trajectory is a relationship between positions in a chosen frame. An observer moving horizontally at constant velocity u measures x′=x−ut if both origins coincide at launch. The vertical coordinate is unchanged because the observer has no vertical motion. Time is shared in this nonrelativistic model.

2. Explain the model

With x=vₓ₀t, the observer measures horizontal velocity vₓ₀−u. It can be positive, zero or negative without changing the physical launch. A ground observer can see a curved path while an observer moving with the projectile’s horizontal component sees a vertical up-and-down path. Neither view changes the downward acceleration.

The observer-frame horizontal displacement at landing is (vₓ₀−u)T. It must not be called the ground range unless u=0. Distinguish the location of the landing point in ground coordinates from the projectile’s separation from the moving observer at that same instant.

The constant-speed observer is an assumption, not a result of the plot. If the observer accelerates, subtracting only ut is wrong. Shared axes also matter: reversing the positive direction changes component signs even for an observer at rest. Record the transformation before comparing measured values.

3. Work an example

Let ground horizontal velocity be 5 m/s, upward velocity 10 m/s and g=10 m/s². Flight time is 2 s. A 3 m/s observer sees horizontal velocity 2 m/s and landing displacement 4 m. The ground range is 10 m. An observer moving at 5 m/s sees zero horizontal displacement throughout.

4. Predict and investigate

Predict first: Can the projectile move backward in one observer’s coordinates while moving forward relative to the ground?

Fix ground horizontal velocity at 5 m/s and compare observer velocities −3,3 and 5 m/s. Then choose a ground horizontal velocity below the observer velocity to make relative motion negative. Check that apex height, return time and vertical landing velocity remain unchanged. Use the lesson experiment controls and retain the settings with your evidence.

5. Transfer and explain

Try a new practice case

A projectile returns to launch height after 3 s with ground horizontal velocity 6 m/s. An observer translates at 2 m/s in that direction. Find the projectile’s observer-frame horizontal displacement.

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

Explain why a changed trajectory shape does not by itself show a changed force. What would distinguish a moving frame from a changed launch?

Record a claim, numerical or graphical evidence, assumptions, a limitation and a changed-case prediction. A supplied synthetic investigation is not evidence that you performed a physical laboratory.

Framework and source notes

Original teaching prose, cases and synthetic datasets. Framework mapping: 1.4.A; 1.4.B; 1.5.B. College Board AP Physics 1 CED, Unit 1, checked 30 October 2026. This mapping is a content cross-reference, not AP authorization or a claim of full course completion.

Apply

A projectile moves horizontally at 4 m/s in the ground frame. An observer moves at 6 m/s in the same direction. What horizontal velocity does that observer measure?

Public practice checks a specific claim. It does not establish independent mastery.

Explain why a changed trajectory shape does not by itself show a changed force. What would distinguish a moving frame from a changed launch?

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

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