Vertical Projectile Motion: Grade 12 Physical Sciences
Throw a ball straight up and it slows, stops for an instant, and comes back down. The acceleration is 9,8 m·s-2 downwards the entire time, including at the top when the ball is not moving at all.
That one sentence is the whole chapter, and it is the thing learners refuse to believe. This page covers free fall, the four equations of motion, sign conventions, the three graphs, bouncing balls, and worked examples in both directions.
The idea everyone argues with
At the highest point the velocity is zero. The acceleration is not.
Learners see "it has stopped" and write a = 0. But acceleration is the rate at which velocity changes, and the velocity is changing faster than ever at that moment: it goes from a small upward value to a small downward value in a fraction of a second.
What would a = 0 at the top actually mean? It would mean the ball stops and stays there. Balls do not hang in the air.
| At the highest point | Value |
|---|---|
| Velocity | 0 m·s-1 |
| Acceleration | 9,8 m·s-2 downwards |
| Net force | The object's weight, downwards |
Free fall, and what counts as it
Free fall is motion in which the only force acting on the object is gravity.
Air resistance has to be ignored. In every school question it is, and the question will say so.
| Term | Means |
|---|---|
| Free fall | Motion of an object where the only force acting is gravity |
| Projectile | An object moving under gravity alone, with a vertical component of motion |
| Trajectory | The path the projectile follows. Straight up and down here, so a vertical line |
A ball thrown upwards is in free fall the moment it leaves your hand, even while it is still going up. Free fall does not mean falling. It means gravity is the only thing acting.
Why the mass does not matter
Newton's Second Law on an object in free fall:
- Fnet = ma
- The only force is weight, so mg = ma
- The mass cancels: a = g
A cricket ball and a marble dropped together land together. They do in a vacuum, and in a classroom they are close enough that you can see it. A sheet of paper does not, which is air resistance, not gravity, and crumpling the paper into a ball proves it.
g is 9,8 m·s-2 and it always points downwards, whether the object is rising, falling or momentarily still.
The sign convention, and why you must write it down
Choose a positive direction before you write anything else, and say which one you chose.
Both choices are correct. What is not correct is changing halfway through.
| If you choose | Then g is | And a ball thrown up has vi |
|---|---|---|
| Upwards positive | −9,8 m·s-2 | Positive |
| Downwards positive | +9,8 m·s-2 | Negative |
The convention that saves the most trouble is upwards positive for anything thrown up, downwards positive for anything dropped. That way the number you are solving for usually comes out positive and the arithmetic is easier to sanity-check.
Write "taking upwards as positive" as the first line of the answer. It is worth a mark in most memos and it stops you contradicting yourself three lines later.
The four equations of motion
These are on the data sheet. For vertical motion, replace a with g and Δx with Δy.
| Equation | Use it when |
|---|---|
| vf = vi + aΔt | You have or want time and are not interested in displacement |
| Δy = viΔt + ½aΔt² | You have or want time and displacement |
| vf² = vi² + 2aΔy | There is no time in the question at all. The most useful one |
| Δy = ((vi + vf) ÷ 2)Δt | You have both velocities and want displacement |
Pick the equation by looking at what is missing, not at what is given. List your five quantities, see which one the question never mentions, and choose the equation that leaves it out.
Two facts that turn hard questions into easy ones
- At the highest point, vf = 0. That is the hidden given in almost every "maximum height" question
- Dropped from rest means vi = 0. The word "dropped" is doing work in the question
Worked example: thrown straight up
A stone is thrown vertically upwards at 12 m·s-1. Find the maximum height and the total time in the air before it returns to the thrower's hand.
Taking upwards as positive, so a = −9,8 m·s-2.
Maximum height. At the top, vf = 0. Time is not mentioned, so use the third equation.
- vf² = vi² + 2aΔy
- 0 = (12)² + 2(−9,8)Δy
- Δy = 144 ÷ 19,6
- Δy = 7,35 m
Time to the top.
- vf = vi + aΔt
- 0 = 12 + (−9,8)Δt
- Δt = 1,22 s
Total time in the air = 2,45 s, because the trip up and the trip down take the same time when it returns to the same height.
That symmetry is worth knowing and worth being careful with. It only holds when the object comes back to the height it started from. The moment it lands lower than it started, the two halves are not equal and you cannot double anything.
Worked example: dropped
A ball is dropped from a balcony 20 m above the ground. Find the time to reach the ground and the speed on impact.
Taking downwards as positive, so a = +9,8 m·s-2 and vi = 0.
- Δy = viΔt + ½aΔt²
- 20 = 0 + ½(9,8)Δt²
- Δt² = 20 ÷ 4,9 = 4,08
- Δt = 2,02 s
For the impact speed, use the equation with no time in it as a check on yourself:
- vf² = vi² + 2aΔy = 0 + 2(9,8)(20) = 392
- vf = 19,8 m·s-1 downwards
Always give the direction. Velocity is a vector and "19,8 m·s-1" on its own is an incomplete answer.
Worked example: thrown up from a height, which is the hard one
A ball is thrown vertically upwards at 8 m·s-1 from the top of a 25 m building. Find the time to hit the ground and the speed on impact.
Taking upwards as positive, so a = −9,8 m·s-2.
The displacement is −25 m, not +25 m and not 33 m. Displacement is final position minus initial position, and the ground is 25 m below the throwing point. The ball goes up first, but that does not change where it finishes.
- Δy = viΔt + ½aΔt²
- −25 = 8Δt + ½(−9,8)Δt²
- 4,9Δt² − 8Δt − 25 = 0
- Δt = (8 ± √(64 + 490)) ÷ 9,8
- Δt = 3,22 s, taking the positive root
The quadratic gives two answers and one of them is negative. Discard it. A negative time is the maths describing where the ball would have been before it was thrown.
Impact speed:
- vf² = (8)² + 2(−9,8)(−25) = 64 + 490 = 554
- vf = 23,5 m·s-1 downwards
Notice it lands faster than the 19,8 m·s-1 of the dropped ball, which makes sense: it had an 8 m·s-1 head start by the time it came back past the throwing point.
The three graphs
Exam papers ask for these more often than they ask for calculations. They are worth drilling.
For a ball thrown up and caught again, taking upwards as positive:
| Graph | Shape | Why |
|---|---|---|
| Position against time | A downward parabola | Rises, flattens at the top, falls symmetrically |
| Velocity against time | A straight line with negative gradient, crossing zero at the top | Velocity changes at a constant rate. The gradient is the acceleration |
| Acceleration against time | A horizontal line at −9,8 | Constant throughout, including at the top |
The velocity-time graph is the one that carries the marks, because it shows three things at once: the gradient is g, the point where it crosses the axis is the highest point, and the area under it is the displacement.
If you choose downwards as positive, every graph flips vertically. The physics is identical. The marker will accept either, so long as your axis label says which you chose.
The bouncing ball
A ball dropped, bouncing several times, is a favourite question.
| Graph | What it looks like |
|---|---|
| Position | A series of parabolic arcs, each one lower than the last |
| Velocity | Sloped straight lines, each ending in a near-vertical jump from negative to positive at each bounce |
| Acceleration | Constant at −9,8 while in the air, with a large positive spike at each bounce |
The bounce is where the marks are. While the ball is touching the ground the floor pushes up on it, so for that instant it is not in free fall and the acceleration is large and upwards. Every arc is lower than the one before because energy is lost in each bounce.
Where this fits in the curriculum
| Subject | Physical Sciences |
|---|---|
| Grade | 12 |
| Term | 1 |
| Topic | Mechanics |
| Status | Examinable theory. No formal assessment attached |
| Paper | Physics, Paper 1 |
It comes straight after momentum and impulse and it is the second Grade 12 mechanics chapter. Nearly every Paper 1 carries a question on it, usually one scenario with a calculation and a graph hanging off it.
The mistakes that cost the marks
| The mistake | What to do instead |
|---|---|
| Writing a = 0 at the highest point | The velocity is zero. The acceleration is 9,8 m·s-2 downwards throughout |
| Not stating the sign convention | First line of every answer: "taking upwards as positive". It is usually a mark |
| Changing the convention halfway through | Pick one and stay in it for the whole question, including the graph |
| Using +25 m when the object lands below where it started | Displacement is final minus initial. Landing 25 m lower is −25 m |
| Doubling the time when the object does not return to its start height | The symmetry only holds for equal heights. Otherwise solve the quadratic |
| Giving a velocity with no direction | It is a vector. "Downwards" or a sign is part of the answer |
| Keeping the negative root of the quadratic | Negative time is before the throw. Discard it |
| Thinking heavier things fall faster | The mass cancels in mg = ma. Only air resistance separates them |
Seeing it without apparatus
You do not need a free-fall timer to make this real.
Drop two objects of very different mass from the same height at the same instant, a textbook and a pen, and let the class listen for one thud or two. They expect two. They hear one.
Then drop a flat sheet of paper against the same textbook. The paper loses badly. Now crumple the paper into a tight ball and repeat. It keeps up. Nothing about its mass changed, so the difference was never mass.
That sequence takes ninety seconds and it settles the argument for the year.
If you want numbers, a ticker tape timer on a falling mass gives a measurable value for g, and the tape itself is a velocity-time graph you can hold in your hand.
How the 40 marks are made up
| Section | Marks |
|---|---|
| Multiple choice | 10 |
| Definitions and the acceleration at the top | 6 |
| Thrown upwards calculation | 10 |
| Dropped object calculation | 8 |
| Graphs | 6 |
No mark allocation is prescribed for this chapter. The split above is ours, weighted towards the calculations and the graphs because that is where Paper 1 puts its marks.
Free worksheet and marking memo
Both free, no sign up, straight to the PDF.
- Learner worksheet, 40 marks, with multiple choice, both directions of calculation, a thrown-from-a-height question and the three graphs
- Marking memorandum, with full working, the mark breakdown line by line and a note on the eight places learners drop marks
Related pages
- Work, energy and power, Grade 12. The same ball, solved with energy instead of equations
- Conservation of momentum, Grade 12. The chapter immediately before this one
- Ticker tape motion investigation. How to measure acceleration for yourself
Apparatus
None of this needs equipment, and that is worth saying plainly. Two objects and a bit of floor will demonstrate the central point.
If you want to measure g rather than assert it, a ticker tape timer is the cheapest honest way to do it, and the tape doubles as a graph.