Thrown straight upwards
Now the opposite case: what happens when an object is thrown vertically upwards?
The problem
An object is sent vertically upwards with a velocity of 30 m s−1. What is the maximum height it rises to? (Take g = 10 m s−2 for simplicity.)
How the velocity changes
| Time t (s) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Velocity v (m s−1) | 30 | 20 | 10 | 0 |
| t (s) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| v (m s⁻¹) | 30 | 20 | 10 | 0 |
At the top, the velocity is zero
After 3 seconds the velocity reaches 0. That is the moment the object arrives at its maximum height — from there it starts to fall back.
So to find "the time taken to reach maximum height", find when the velocity becomes zero.
The maximum height
This is the area of the triangle under the graph.
(30 m s−1 ÷ 2) × 3 s = 45 m
Why does the graph slope downwards?
In drawing this graph, velocity vertically upwards has been taken as positive. That is why the acceleration due to gravity appears on it as a negative acceleration.
Chapter summary
Chapter summary
- The distance travelled depends on the route taken. Displacement depends only on the starting and finishing positions.
- Distance has only a magnitude — it is a scalar quantity. Displacement also has a direction — it is a vector quantity.
- Speed = distance ÷ time (a scalar). Velocity = displacement ÷ time (a vector).
- Acceleration = change in velocity ÷ time. A negative acceleration is a deceleration. Both are vector quantities.
- On a displacement–time graph the gradient gives the velocity. On a velocity–time graph the gradient gives the acceleration and the area gives the displacement.
Glossary
| Distance | දුර |
| Displacement | විස්ථාපනය |
| Object | වස්තුව |
| Vector quantity | දෛශික රාශිය |
| Scalar quantity | අදිශ රාශිය |
| Speed | වේගය |
| Velocity | ප්රවේගය |
| Acceleration | ත්වරණය |
| Retardation / Deceleration | මන්දනය |
| Acceleration due to gravity | ගුරුත්වජ ත්වරණය |