Lessons · freefall
A ball thrown straight up
Regime 1 — the algebra is the calculus, evaluated. Step the algebra, step the calculus, and watch the algebra formula fall out of the integral. SymPy proves the two registers agree.
A ball leaves your hand at 20 m/s straight up, from a height of 1.5 m. How high does it go, when does it land, and how fast is it moving when it hits the ground?
Constant — the only force acting is gravity. On the stacked graph this is a flat line.
simplify(algebra - calculus) == 0 for each identity below
- ✓ The algebra formula is the integral with . [structural]
- ✓ The algebra formula is the integral with . [structural]
- ✓ Velocity is zero at the apex time — so the apex time is correct. [structural]
- ✓ Max height from the timeless equation equals the calculus trajectory at its apex. [structural]
- ✓ At the flight time the calculus trajectory returns to the ground — so the flight time is correct. [structural]
- ✓ Impact velocity from the timeless equation equals the calculus velocity at landing. [structural]
Dimensional homogeneity: checked by SymPy (holds).
On the v-t graph the velocity passes through zero at the apex, but the line never bends: its slope is m/s² the whole way. for an instant; is never zero.
Modeling assumptions — author-asserted, disclosed not discharged
- No air resistance (the ball is in free fall under gravity alone).
- g = -10 m/s^2, a clean-arithmetic simplification of -9.81 m/s^2.
- The ball is a point mass; its size and spin are ignored.
The stacked graph, fully annotated
A static rendering (Matplotlib) at the default parameters — the interactive version is in the Graph tab above.
Formulas used
Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):
- Position under constant acceleration
Valid when: acceleration is constant
Open in reference →
- Velocity under constant acceleration
Valid when: acceleration is constant
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- Constant acceleration
Valid when: acceleration is constant
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