Choosing a sign convention
Choose up (or down) as positive and keep it for the whole question. Taking up as positive: the acceleration due to gravity is −9,8 m·s⁻² at all times while the ball is in the air, whether it is moving up, at the top, or moving down.
What each graph shows
| Graph | Gradient gives | Area gives | Shape for free fall |
|---|
| Position–time | Velocity | — | Parabola (curve) |
| Velocity–time | Acceleration | Displacement | Straight line, gradient −9,8 m·s⁻² |
| Acceleration–time | — | Change in velocity | Horizontal line at −9,8 m·s⁻² |
A ball that bounces
A ball dropped onto the floor (up positive):
- Falling: velocity becomes more negative along a straight line of gradient −9,8; position decreases along a curve.
- At the bounce: velocity changes from negative to positive almost instantly — on the v–t graph this is a nearly vertical jump. The rebound speed is smaller than the impact speed if the bounce is inelastic.
- Rising: velocity decreases from its positive value to zero at the top, along a line with the same gradient (−9,8) as before — the lines are parallel.
- Acceleration–time: a horizontal line at −9,8 m·s⁻², interrupted by a short, very large positive spike during contact with the floor, when the floor exerts a large upward force.
Worked example: a dropped ball
✏️ Ball dropped from 1,8 m, rebounds at 4,0 m·s⁻¹
1Speed at impact: vf² = vi² + 2aΔy = 0 + 2(9,8)(1,8) → vf = 5,94 m·s⁻¹ downwards.
2Time to fall: Δt = Δv/a = 5,94 ÷ 9,8 = 0,61 s.
3Height after the bounce: 0 = 4,0² − 2(9,8)Δy → Δy = 0,82 m.
4The rebound speed (4,0 m·s⁻¹) is less than the impact speed (5,94 m·s⁻¹), so kinetic energy was lost: the bounce is inelastic.
Graph-sketching checklist: label both axes with quantities and units; mark known values (times, velocities); draw parallel lines for the free-fall sections; show the velocity jumping from negative to positive at the bounce; make the second peak lower if the bounce is inelastic.