Rubber Ball Drop and Rebound Simulation

Compare velocity-time and speed-time graphs. Downward is positive for velocity, but speed is always positive.

Ball Motion

Release point: 20 m above ground
New maximum height: 5 m
Ground: 0 m
Time 0.00 s
Height above ground 20.00 m
Velocity 0.00 m/s
Speed 0.00 m/s
Stage Released
0.00 s

Graph Comparison

Velocity-Time Graph

Speed-Time Graph

Main teaching point

Velocity includes direction. Since downward is positive, upward motion has negative velocity.

Speed does not include direction. It is the size of the velocity only: speed = |velocity|.

Question 13 Answers

Time Situation Velocity Speed
0 s Released from rest 0 m/s 0 m/s
2 s Just before hitting ground +20 m/s 20 m/s
2 s Just after rebound -10 m/s 10 m/s
3 s At new maximum height 0 m/s 0 m/s

(a) Sketch the speed-time graph

The speed-time graph goes from:

(0, 0) to (2, 20), then drops suddenly to (2, 10), then decreases to (3, 0).

The graph does not go below the time axis because speed is always positive.

(b) Speed V just before first hit

v² = u² + 2as
v² = 0² + 2(10)(20)
v² = 400
v = 20 m/s

Answer: V = 20 m/s

(c) Total distance travelled

The ball first falls 20 m.

After rebound, its upward speed is half of 20 m/s: 10 m/s.

Rebound height: 0² = 10² - 2(10)s

100 = 20s, so s = 5 m.

Total distance travelled: 20 + 5 = 25 m

Answer: 25 m

(d) Change during rebound

Just before rebound, velocity is: +20 m/s

Just after rebound, velocity is: -10 m/s

Change in velocity: -10 - (+20) = -30 m/s

So the physically correct change in velocity during rebound is: -30 m/s.

If the question intended change in speed, then the speed changes from 20 m/s to 10 m/s, so:

10 - 20 = -10 m/s

This explains why the answer key may show -10 m/s: it is treating the quantity as change in speed, not change in velocity.