Module 1 · motion in two dimensions
Almost everything in two-dimensional motion — the parabola, the monkey and the hunter, the banked road, the leaning motorbike — comes out of a single idea that takes one line to write and a lifetime to stop being surprised by.
Two identical steel balls sit at the same height. At the same instant, one is simply released, and the other is fired horizontally out of a spring launcher at 3.5 m/s. No air resistance yet.
Commit before you look. Which ball reaches the floor first?
There is no penalty for being wrong. There is a real penalty for skipping the guess — you learn far less from watching something you never took a position on.
Every rung is horizontal, at every speed you can set, always. Slide the launch speed to 8 m/s and the fired ball flies right across the bench — and still touches down on the same beat as the ball that just fell.
Write the position of a ball as a vector, and Newton's second law as a vector equation. Near the Earth's surface the only force is gravity, which points straight down:
Now split that single vector equation into its components. A vector equation is not one statement — it is two statements stacked, one per direction, and they do not talk to each other:
The landing time comes only from the vertical line, by setting y = 0. The horizontal line is not consulted, is not asked for permission, and has no vote. That is the entire explanation.
Say it as a sentence you can carry: time is shared, directions are not. The clock is common to both components. Everything else is private.
That independence of the two directions has a name — decoupling, because the two equations are not coupled to each other. It is not a law of the universe. It is a consequence of gravity having no horizontal component. Add air resistance and the picture collapses immediately, because drag points opposite to the total velocity and has size roughly proportional to v² = vx² + vy². Now the horizontal equation contains vy and the vertical equation contains vx: they are coupled, no clean formula survives, and you solve it numerically. This is why gunners fire ranging shots and print the results as a table of numbers instead of using a formula, and why working out how far a cricket ball will actually carry is a genuinely hard calculation.
A monkey hangs from a branch. A zoologist below has a tranquilliser gun. The monkey is sharp: it has learned to let go of the branch the instant it sees the flash at the end of the barrel. It is quite sure this drops it out of the way.
Where should she aim?
Tick Show the no-gravity world and fire again. Two pale ghosts appear: where the dart would be if gravity switched off, and where the monkey would be. The ghost dart travels in a perfectly straight line and meets the ghost monkey exactly where it was hanging, because that is what aiming straight at something means.
Take the real position of each object and subtract the position it would have had with gravity switched off. For any object starting anywhere with any velocity:
Both the dart and the monkey sag downward by exactly ½gt² from where they would otherwise have been. If two ghosts meet at some time t, then their real versions meet at that same time, just lower down by the same amount. Aiming straight at the target is therefore always correct — and the dart hits the monkey somewhere below the branch.
The one condition is that the meeting must happen before the ground gets there. Turn the muzzle speed down to 7 m/s and you will see the dart is still correct in principle and useless in practice: both of them land first.
You just did the whole problem in a freely falling frame of reference — a coordinate system that itself falls at g. In that frame gravity vanishes, and both objects move in straight lines at constant velocity. Straight lines are easy; parabolas are not. Walter Lewin builds his entire lecture 4 around this move, and it is the same idea Einstein started from in 1907 when he called it "the happiest thought of my life" — a falling person does not feel their own weight. From that seed came general relativity.
Until now the launch has been horizontal. Now launch at speed u at some tilt above the horizontal — call it θ, the launch angle. A launch velocity is a vector like any other, so it splits into components the same way: u cos θ across, u sin θ up. Feed those into the two lines from Bench 1 (taking the launch point as y = 0):
The horizontal line gives t = x/(u cos θ); substitute that into the vertical line and the shape of the path falls out:
For a launch and landing at the same height, put y = 0 in the vertical line: 0 = (u sin θ)t − ½gt². Factorise: t(u sin θ − ½gt) = 0. Either t = 0, which is the launch itself, or t = T = 2u sin θ/g, which is the landing. The path is symmetric, so the ball is at its highest at T/2; put that back into the vertical line and out drops H. Put the whole of T into the horizontal line (x = u cos θ · t) and out drops R.
Because sin 2θ peaks at 2θ = 90°, the flat-ground range is largest at θ = 45°. And because sin 2θ = sin (180° − 2θ), any two angles adding to 90° — 30° and 60°, 20° and 70° — give the same range by completely different-looking paths. That is not a coincidence. Swap 30° for 60° and you swap over how much of the launch speed goes forward and how much goes upward — one buys range directly, the other buys the time to use it, and the range depends on the two multiplied together. The sine identity is the algebra noticing what the physics was already doing.
A shot-putter releases the shot from about 2 m above the ground, at roughly 13 m/s. To throw as far as possible, she should release at:
Start with the launch height at zero and confirm the textbook: the peak of the lower curve sits exactly at 45°, and the complementary path lands on the same spot. Now raise the launch height. The peak slides left, and keeps sliding — and the two complementary paths no longer agree.
With a launch height h, solving y = 0 needs the full quadratic formula rather than the tidy root:
Set dR/dθ = 0 and, after some patient algebra, the optimum angle satisfies
Look at how h enters the answer: only ever in the combination gh/u². That combination is a pure number, because it is one height divided by another. The second height is hiding — throw straight up at speed u and you rise u²/2g, so u²/g is a height the launch speed alone can reach. The question is therefore never "is 2 m a lot?" It is always "is 2 m a lot compared with the height this launch speed could climb to on its own?" A 2 m release matters enormously for a 13 m/s shot put (which could climb about 8.6 m) and not at all for a 900 m/s bullet (which could climb over 40 km). Getting into the habit of asking "compared to what?" is most of what separates physics from formula-substitution.
Real shot-putters release at about 37° — lower even than this model predicts. The reason is the body. Pushing steeply upward you are working against your own weight the whole way, so you simply cannot get the shot moving as fast as you can at a shallower angle. In other words u is not a fixed number handed to you: it drops as θ rises. A real optimisation has to allow for that too. Long-jumpers take off near 20° for the same reason, pushed much further. Sport is where clean physics meets a body that has its own opinions.
Bench 1 did start from a force — Newton's second law was the first thing we wrote down. But we used it once, to get the acceleration, and then spent the rest of the section describing the motion that followed. Describing motion, given the acceleration, is kinematics. Dynamics runs the other way: you look at the situation, work out which forces are actually acting, and let F = ma tell you what the acceleration must be. Making that switch costs you nothing new, because F = ma is a vector equation too, and it splits into components exactly the same way.
Here is the case where that pays off. A car goes round a bend of radius r at speed v. It is not speeding up or slowing down, yet it is accelerating, because velocity is a vector and its direction is changing. That acceleration points at the centre of the circle and has size v²/r — a result derived properly in Module 6; take it on trust for now. Something must supply it.
One piece of vocabulary before you commit. A solid surface can only push outward along its own perpendicular: it cannot pull you towards it, and it cannot push you sideways along itself — that sideways job belongs to friction. The perpendicular push is called the normal force, written N; "normal" here is the geometry word for "at right angles", not "unremarkable".
On a properly banked bend taken at exactly the design speed, what supplies the centre-pointing force?
Set μ, the coefficient of friction between tyre and road, to 0 — a sheet of wet ice — and hunt for the one speed at which the car holds its line. Now draw the car with every force on it and nothing else on the page: its weight mg straight down, and the road's normal push N at right angles to the tilted surface. A drawing made this way is called a free-body diagram; Module 3 makes a discipline of it. Choose axes next. Horizontal and vertical, not along the slope, because the acceleration is horizontal — one axis, one job — and then read the two components off the diagram:
Divide the first by the second. The normal force cancels, and so does the mass:
Now put friction back. It can act up the slope (stopping a slow car sliding in) or down the slope (stopping a fast car sliding out), which turns the single design speed into a band:
Watch the denominator of vmax. As μ tan θ approaches 1, the predicted maximum speed runs away to infinity — which is nonsense, and is the model announcing its own limits: long before that, the car tips over instead of sliding (whether it tips first or slides first is set by the ratio of its track width to its centre-of-mass height, which is Module 10's business, not ours), and at high enough banking it would need the tyres to hold it on like Velcro. A formula that blows up is telling you something real about where you left physics behind.
Highway bends in India are banked to less than the frictionless tan θ = v²/rg demands — IRC practice caps superelevation at about 7% (roughly 4°) so a vehicle stopped on the bend in the wet does not slide inwards; tyre friction picks up the rest. Which is why the posted advisory speed on a flyover ramp is a physics statement, not a suggestion. A velodrome banks past 40° because riders need the design speed high. And a motorcyclist leaning into a corner is running exactly the same equation with the bank angle built out of the bike's own tilt instead of the road's — which is why the lean angle depends on speed and corner radius, and not at all on how heavy the rider is.
Attempt each before revealing. Getting one wrong here is worth more than getting all three right by reading ahead.
One idea we have only touched, and it will come back with force in Module 19: frames of reference. When you shift from watching a motion from the ground to riding along with it, you are doing arithmetic — the velocity of A seen by C equals the velocity of A seen by B plus the velocity of B seen by C. Fire a bullet forward from a moving train and the ground observer sees bullet speed plus train speed, by exactly that rule. This is called Galilean velocity addition, and question 3 below turns on it.
1. A ball is thrown at 20 m/s at 30° above the horizontal. At the very top of its flight, what is its acceleration?
2. Two stones are thrown from the same rooftop with the same speed — one at 30° above the horizontal, the other at 30° below. Ignoring air resistance, which hits the ground faster (greater landing speed)?
3. You are cycling north at 5 m/s. Rain is falling vertically at 5 m/s. At what angle from the vertical must you tilt your umbrella?
A fielder on the boundary at Chepauk has to judge a skier — a ball hit steeply up into the air — in a strong crosswind. He has under four seconds. He has never solved a differential equation and never will, and yet he will usually be standing in the right place. What is he doing instead of the calculation? Is that a worse method than the one you learned today, or a different one, or the same one wearing different clothes? Write down a paragraph now — while the question is fresh — and hold on to it. The calculation that would justify what he does is precisely what the clean formulas above cannot deliver, and it is why sport keeps its own body of unwritten physics.
These are the real thing, all free. Watch or read them alongside the modules rather than instead of them — the benches above are for getting your hands dirty; these are for hearing a master phrase it.
Module 1 was deliberately one idea, seen four times: two directions, one law, no exceptions. The other twenty-four modules build out from exactly that habit of mind — through gas laws and energy, circles and waves, fields and circuits, and on into relativity and the quantum, each one roughly 45 minutes. See the full course map for what each module covers and how they connect.