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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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 experimentPredict
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.
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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.
| Quantity | Value | Unit |
|---|---|---|
| Return time | 2 | s |
| Apex time | 1 | s |
| Maximum height | 5 | m |
| Ground horizontal displacement | 10 | m |
| Observer horizontal displacement | 4 | m |
| Horizontal velocity in displayed frame | 2 | m/s |
| Vertical velocity at return | -10 | m/s |
Accessible plot coordinates
| horizontal position (m) | height (m) |
|---|---|
| 0 | 0 |
| 0.2 | 0.95 |
| 0.4 | 1.8 |
| 0.6 | 2.55 |
| 0.8 | 3.2 |
| 1 | 3.75 |
| 1.2 | 4.2 |
| 1.4 | 4.55 |
| 1.6 | 4.8 |
| 1.8 | 4.95 |
| 2 | 5 |
| 2.2 | 4.95 |
| 2.4 | 4.8 |
| 2.6 | 4.55 |
| 2.8 | 4.2 |
| 3 | 3.75 |
| 3.2 | 3.2 |
| 3.4 | 2.55 |
| 3.6 | 1.8 |
| 3.8 | 0.95 |
| 4 | 0 |
| horizontal position (m) | height (m) |
|---|---|
| 0 | 0 |
| 0.5 | 0.95 |
| 1 | 1.8 |
| 1.5 | 2.55 |
| 2 | 3.2 |
| 2.5 | 3.75 |
| 3 | 4.2 |
| 3.5 | 4.55 |
| 4 | 4.8 |
| 4.5 | 4.95 |
| 5 | 5 |
| 5.5 | 4.95 |
| 6 | 4.8 |
| 6.5 | 4.55 |
| 7 | 4.2 |
| 7.5 | 3.75 |
| 8 | 3.2 |
| 8.5 | 2.55 |
| 9 | 1.8 |
| 9.5 | 0.95 |
| 10 | 0 |
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.
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Explain why a changed trajectory shape does not by itself show a changed force. What would distinguish a moving frame from a changed launch?
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