AP · AP C:M · extended response
A moving receiver below a launch platform
AP Physics C: Mechanics · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Receiver position after launch
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Receiver horizontal position / m
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Receiver horizontal position / m |
|---|---|
| 0 | 4 |
| 0.5 | 5 |
| 1 | 6 |
| 1.5 | 7 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Choose your AP course · School year 2026–27 · May 2027 exam
Choose a course to check this question’s reviewed parts and required depth. Its original AP label and topic codes do not establish suitability for the current syllabus.
Choose reviewed AP practice →A ball leaves a platform horizontally at 6.0 m s⁻¹, 5.0 m above the ground. A receiver starts 4.0 m horizontally from the point directly below the launch and moves away at constant speed 2.0 m s⁻¹. Use g = 10 m s⁻² and neglect air resistance; model both objects as points. Supplied data (Time / s; Receiver horizontal position / m): (0, 4), (0.5, 5), (1, 6), (1.5, 7).
- (a)
Determine whether the ball meets the receiver at ground level and calculate the impact velocity components.
3 marks - (b)
The ball is instead launched horizontally at 7.0 m s⁻¹. Find the receiver's required starting position and explain the role of the two independent components.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Determine whether the ball meets the receiver at ground level and calculate the impact velocity components.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Vertical motion gives 5.0 = ½(10)t², hence t = 1.0 s.
- 2
The ball reaches x = 6.0t = 6.0 m; the receiver reaches x = 4.0+2.0t = 6.0 m, so they coincide.
- 3
Horizontal velocity stays 6.0 m s⁻¹; vertical velocity is −gt = −10 m s⁻¹. Speed is √136 ≈11.7 m s⁻¹, directed about 59° below horizontal.
(b)
The ball is instead launched horizontally at 7.0 m s⁻¹. Find the receiver's required starting position and explain the role of the two independent components.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Flight time remains 1.0 s because initial vertical velocity and vertical drop are unchanged.
- 2
The ball reaches x = 7.0 m, while the receiver travels 2.0 m during the fall. Its initial x must therefore be 5.0 m.
- 3
Horizontal speed changes range without changing fall time under this model; a finite receiver or air resistance would require a more detailed tolerance model.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- Vertical motion gives 5.0 = ½(10)t², hence t = 1.0 s.
- The ball reaches x = 6.0t = 6.0 m; the receiver reaches x = 4.0+2.0t = 6.0 m, so they coincide.
- Horizontal velocity stays 6.0 m s⁻¹; vertical velocity is −gt = −10 m s⁻¹. Speed is √136 ≈11.7 m s⁻¹, directed about 59° below horizontal.
- Flight time remains 1.0 s because initial vertical velocity and vertical drop are unchanged.
- The ball reaches x = 7.0 m, while the receiver travels 2.0 m during the fall. Its initial x must therefore be 5.0 m.
- Horizontal speed changes range without changing fall time under this model; a finite receiver or air resistance would require a more detailed tolerance model.
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