Appearance
1.3 — Motion in a Plane
Fire a bullet horizontally from a rifle held one metre above flat ground. At the same instant, drop a second bullet from the same height. Which lands first?
They land together. The fired bullet travels several hundred metres downrange while it falls, and the dropped one goes nowhere sideways, but both are in the air for the same 0.45 seconds. That result is the entire content of this chapter, and it comes from one idea: motion in two directions is two independent one-dimensional motions happening at the same time, and neither one knows the other exists.
To use that idea you need a tool for handling two directions at once without them getting tangled. That tool is the vector.
A vector is a quantity that needs a direction to be complete
Some quantities are fully described by a number and a unit. A temperature of 30 °C, a mass of 5 kg, a time of 12 s — there is no such thing as "30 °C to the north". These are scalars.
Others are not complete until you say which way. A displacement of 5 m is useless without a direction. So is a velocity, a force, an acceleration. These are vectors.
The test is simple and it is about addition, not about arrows. Walk 3 m east, then 4 m north. Your total distance walked is 7 m — the scalars added straight. Your displacement is 5 m north-east, because the two legs of the journey are at right angles and \sqrt{3^2+4^2} = 5. Vectors add by a rule that takes direction into account; scalars just add. That rule is what makes them a different kind of object.
Notation: a vector gets a bold letter \mathbf{v} or an arrow \vec{v}. Its magnitude — its size, stripped of direction, always positive — is written |\vec{v}| or just v in italics. Watch that distinction, because \vec{v} and v mean different things and the difference is exactly the direction.
Adding vectors: the triangle and the parallelogram
To add \vec{A} and \vec{B}, slide \vec{B} (without rotating it) so its tail sits on the head of \vec{A}. The sum \vec{A} + \vec{B} is the arrow from the tail of \vec{A} to the head of \vec{B}. That is the triangle rule, and it is just a picture of "do this journey, then that one".
Draw both vectors from the same starting point instead and complete the parallelogram, and the diagonal is the same sum. The parallelogram rule and the triangle rule are the same rule drawn two ways, and which one you use is a matter of which picture is less cluttered for the problem in front of you.
Subtracting is adding the reverse: \vec{A} - \vec{B} = \vec{A} + (-\vec{B}), where -\vec{B} is \vec{B} turned through 180°. This looks like a technicality and it is about to become the most useful operation in the chapter, because relative velocity is a subtraction.
If \vec{A} and \vec{B} have magnitudes A and B with angle \theta between them, the magnitude of the sum is
|\vec{A}+\vec{B}| = \sqrt{A^2 + B^2 + 2AB\cos\theta}
which is the cosine rule from Volume II, Chapter 3.5, with the sign flipped because the angle inside the triangle is the supplement of \theta. Two checks make it memorable. If \theta = 0 the vectors point the same way, \cos 0 = 1, and the expression collapses to \sqrt{A^2+2AB+B^2} = A+B — they add straight, as they must. If \theta = 180°, \cos 180° = -1, and it collapses to |A - B| — they cancel as far as they can. So the sum of two vectors always lies between |A-B| and A+B, and where in that range depends entirely on the angle.
Components: the move that makes vectors easy
Working with magnitudes and angles is painful. Working with components is not, and it is what you will actually do.
Any vector in a plane can be written as the sum of one piece along the x-axis and one along the y-axis. If \vec{A} has magnitude A and makes an angle \theta with the x-axis, drop a perpendicular from its head to each axis and read off:
A_x = A\cos\theta, \qquad A_y = A\sin\theta
Cosine goes with the axis the angle is measured from; sine goes with the other one. That is the only thing to remember, and if you ever doubt it, set \theta = 0: the vector lies along x, so A_x should be all of it and A_y should be nothing, and indeed \cos 0 = 1 and \sin 0 = 0.
Going back the other way:
A = \sqrt{A_x^2 + A_y^2}, \qquad \theta = \tan^{-1}\!\left(\frac{A_y}{A_x}\right)
The magnitude is Pythagoras, because the components are at right angles by construction. (When using the inverse tangent, check which quadrant you are in — the calculator cannot tell (-3, -4) from (3, 4) and will happily give you an angle pointing the wrong way.)
The reason components matter so much: once everything is in components, vector addition becomes ordinary arithmetic done twice.
\vec{C} = \vec{A}+\vec{B} \quad\Longleftrightarrow\quad C_x = A_x + B_x \ \text{ and } \ C_y = A_y + B_y
No triangles, no cosine rule, no angles. Break every vector into x and y, add the numbers in each column, and reassemble at the end if you need a magnitude. Every problem in the rest of this Part is done this way.
Motion in a plane, and the independence of the two directions
Position in a plane is a vector, \vec{r} = (x, y). Velocity and acceleration are its derivatives, taken component by component:
\vec{v} = \frac{d\vec{r}}{dt} = \left(\frac{dx}{dt}, \frac{dy}{dt}\right) = (v_x, v_y), \qquad \vec{a} = \frac{d\vec{v}}{dt} = (a_x, a_y)
And here is the crucial structural fact, which deserves to be stated on its own:
The x-equation contains only x-quantities. The y-equation contains only y-quantities. They never mix.
That is why the two bullets land together. Gravity acts downwards, so it appears only in the y-equation. The fired bullet's large horizontal velocity lives entirely in the x-equation, where gravity has no say. The vertical motion of both bullets is identical — start at rest vertically, accelerate down at g — so they take identical time to fall. The horizontal motion changes where they land, never when.
This independence is not obvious and it was not obvious historically. It is Galileo's again, from 1638, and it is the reason cannon range tables could finally be calculated instead of guessed.
Projectile motion, derived in full
A projectile is anything launched and then left alone, with gravity the only force acting. We ignore air resistance for now; the last section says exactly how much of a lie that is.
Set up: launch from the origin at speed u, at angle \theta above the horizontal. Take up and right as positive.
Split the launch velocity into components:
u_x = u\cos\theta, \qquad u_y = u\sin\theta
Now the accelerations. Horizontally there is no force at all, so
a_x = 0
Vertically, gravity pulls down, so
a_y = -g
That pair of lines is the whole of projectile motion. Everything below is those two facts pushed through the equations of Chapter 1.2, once for each direction.
Horizontal motion
With a_x = 0, the first equation of motion gives v_x = u_x + 0\cdot t, so:
v_x = u\cos\theta \quad \text{(constant, for the entire flight)}
and the second gives:
x = u\cos\theta \cdot t
The horizontal velocity never changes. Not at the top, not on the way down, never. Nothing is pushing or pulling sideways, so by Newton's first law nothing changes sideways.
Vertical motion
With a_y = -g:
v_y = u\sin\theta - gt
y = u\sin\theta\cdot t - \tfrac12 g t^2
This is exactly the thrown-ball problem from Chapter 1.2. The vertical part of a projectile is a ball thrown straight up at u\sin\theta; the horizontal part is a puck sliding on frictionless ice at u\cos\theta. Nothing new is happening.
Time to the top
At the highest point the projectile is momentarily neither rising nor falling, so v_y = 0:
0 = u\sin\theta - g\,t_{\text{top}} \quad\Longrightarrow\quad t_{\text{top}} = \frac{u\sin\theta}{g}
Note it is v_y that vanishes, not \vec{v}. At the top the projectile is still moving, at speed u\cos\theta, horizontally. This catches people constantly: "at the top the velocity is zero" is true for a vertical throw and false for every other launch angle.
Maximum height
Use v^2 = u^2 + 2as in the vertical direction, with final v_y = 0 and a = -g:
0 = (u\sin\theta)^2 - 2gH \quad\Longrightarrow\quad \boxed{H = \frac{u^2\sin^2\theta}{2g}}
Total time of flight
The projectile lands when it returns to y = 0:
0 = u\sin\theta\cdot T - \tfrac12 gT^2 = T\left(u\sin\theta - \tfrac12 gT\right)
Discard the T = 0 root (that is the launch), and:
\boxed{T = \frac{2u\sin\theta}{g}} = 2\,t_{\text{top}}
Up and down take exactly the same time. That symmetry holds only because the launch and landing heights are equal and there is no air resistance; break either condition and it fails.
Range
Range R is the horizontal distance covered in the full flight time. Horizontal velocity is constant, so it is simply speed times time:
R = u\cos\theta \times T = u\cos\theta \times \frac{2u\sin\theta}{g} = \frac{2u^2\sin\theta\cos\theta}{g}
Now use the double-angle identity 2\sin\theta\cos\theta = \sin 2\theta (Volume II, Chapter 3.5):
\boxed{R = \frac{u^2\sin 2\theta}{g}}
That compact form pays for itself immediately.
Maximum range is at 45°. R is largest when \sin 2\theta is largest, and sine peaks at 1 when its argument is 90°. So 2\theta = 90°, giving \theta = 45°, and R_{\max} = u^2/g.
Complementary angles give the same range. \sin 2\theta = \sin(180° - 2\theta), and 180° - 2\theta = 2(90° - \theta). So launching at \theta and at (90° - \theta) lands you in exactly the same place. A shell fired at 30° and one fired at 60° hit the same target — the 60° shot goes much higher and takes much longer, but lands on the same spot. Artillery calls these the low and high trajectories, and the high one is chosen when you need to clear a ridge.
The shape of the path is a parabola
We have x and y each as a function of t. Eliminate t to get the shape of the curve directly. From the horizontal equation:
t = \frac{x}{u\cos\theta}
Substitute into the vertical equation:
y = u\sin\theta\left(\frac{x}{u\cos\theta}\right) - \frac{g}{2}\left(\frac{x}{u\cos\theta}\right)^2
The first term simplifies because \sin\theta/\cos\theta = \tan\theta and the u's cancel:
\boxed{y = x\tan\theta - \frac{g\,x^2}{2u^2\cos^2\theta}}
Look at the structure: this is y = (\text{constant})x - (\text{constant})x^2, which is a quadratic in x with a negative leading coefficient. That is the equation of a downward-opening parabola (Volume II, Chapter 2.3). Every projectile in vacuum flies a parabola, and this is the proof — not a resemblance, an identity.
Worked example: the ball off the table
A ball rolls off a table 1.20 m high at 2.5\ \text{m/s} horizontally. Where does it land, and how fast is it going when it does?
Rolling off horizontally means \theta = 0, so u_x = 2.5\ \text{m/s} and u_y = 0.
Time of fall comes entirely from the vertical motion, which knows nothing about the 2.5 m/s. Taking down as positive for this part:
1.20 = 0\cdot t + \tfrac12(9.81)t^2 \quad\Longrightarrow\quad t^2 = \frac{2.40}{9.81} = 0.2447 \quad\Longrightarrow\quad t = 0.495\ \text{s}
Horizontal distance:
x = 2.5 \times 0.495 = 1.24\ \text{m}
Landing velocity. Horizontally it is still 2.5 m/s. Vertically:
v_y = 0 + 9.81\times0.495 = 4.86\ \text{m/s}
The two are perpendicular, so the speed is Pythagoras:
v = \sqrt{2.5^2 + 4.86^2} = \sqrt{6.25 + 23.6} = \sqrt{29.9} = 5.47\ \text{m/s}
at an angle below the horizontal of
\tan^{-1}\!\left(\frac{4.86}{2.5}\right) = 62.8°
Note that the ball's speed increased from 2.5 to 5.47 m/s even though nothing pushed it forwards. Gravity added vertical speed, and the total is the vector combination.
Relative velocity, and why the rain slants
Velocity is always measured with respect to something, and usually the something is so obviously the ground that nobody mentions it. When two things are both moving, you have to be explicit.
Write \vec{v}_{AB} for "the velocity of A as measured by B". The rule that generates every result in this section is:
\vec{v}_{AB} = \vec{v}_A - \vec{v}_B
Read aloud: the velocity of A relative to B is the velocity of A minus the velocity of B, both measured relative to the ground. The reasoning is straightforward — to see the world as B sees it, subtract B's own motion from everything, which is the same as B standing still and everything else shifting by -\vec{v}_B.
Two cars on a motorway both doing 100 km/h in the same direction: \vec{v}_{AB} = 100 - 100 = 0. They appear stationary to each other, which is why overtaking at speed feels so slow. Head-on, one is -100, so \vec{v}_{AB} = 100 - (-100) = 200 km/h, and that is why head-on collisions are catastrophic.
Crossing a river
You can row at 3\ \text{m/s} in still water. The river flows at 2\ \text{m/s}. The river is 60 m wide.
Case 1: point straight across. Your velocity relative to the water is 3 m/s across; the water's velocity relative to the ground is 2 m/s downstream. Your velocity relative to the ground is the vector sum, and the two are perpendicular:
v = \sqrt{3^2+2^2} = 3.61\ \text{m/s}
The crossing time depends only on the across-component, because only that component reduces the distance to the far bank:
t = \frac{60}{3} = 20\ \text{s}
During those 20 s the current carries you 2 \times 20 = 40\ \text{m} downstream. You arrive at the far bank 40 m below where you aimed. This is the fastest crossing, and it stays the fastest no matter how strong the current gets, because the current contributes nothing at all to the across-direction.
Case 2: land directly opposite. Now you must aim upstream so that the upstream component of your rowing exactly cancels the current. If you row at angle \alpha upstream from straight-across, the along-river component of your velocity is 3\sin\alpha, and cancelling the current needs
3\sin\alpha = 2 \quad\Longrightarrow\quad \sin\alpha = 0.667 \quad\Longrightarrow\quad \alpha = 41.8°
Your across-river component is now only 3\cos(41.8°) = 2.24\ \text{m/s}, so the crossing takes
t = \frac{60}{2.24} = 26.8\ \text{s}
Longer, as it must be — you spent some of your effort fighting the current instead of crossing. And note that if the river flowed faster than you can row, this case has no solution at all: \sin\alpha would have to exceed 1. You simply cannot land directly opposite, and the best you can do is minimise how far downstream you end up.
Rain on a moving train
Rain falls vertically at 5\ \text{m/s}. You are on a train doing 10\ \text{m/s}. Which way should you tilt your umbrella?
Apply the rule. The rain's velocity relative to you is
\vec{v}_{\text{rain,you}} = \vec{v}_{\text{rain}} - \vec{v}_{\text{you}} = (0, -5) - (10, 0) = (-10, -5)
So in your frame the rain comes down and backwards at
\sqrt{10^2+5^2} = 11.2\ \text{m/s}
tilted from the vertical by
\tan^{-1}\!\left(\frac{10}{5}\right) = 63.4°
Tilt the umbrella 63.4° forward from vertical — into your direction of travel. This is why rain streaks slant backwards on a car window even when it is falling dead straight, and why the streaks get more slanted the faster you drive.
What air resistance actually does to all this
Every result above assumed vacuum. Air resistance is a real force, roughly proportional to v^2 at everyday speeds (Chapter 1.11 derives why), always pointing opposite to the motion. Adding it changes the picture in four ways worth knowing now, because otherwise the parabola becomes a superstition.
The path is no longer a parabola. It becomes steeper on the way down than on the way up, because the projectile has lost speed. The shape has a name, the ballistic curve, and it has no simple closed form — real gunnery uses numerical integration.
Range drops, sometimes enormously. A rifle bullet fired at 45° in vacuum would travel about 80 km. In air it manages about 4 km. Air resistance is not a correction to that problem; it is most of the problem.
The optimum angle is no longer 45°. For a fast projectile losing energy to drag, staying in the air longer costs more than it gains, so the best angle drops — around 30–35° for a shot put, lower still for a bullet.
Time up no longer equals time down. The upward leg fights gravity and drag; the downward leg fights only drag. The descent is the longer half.
None of this makes the vacuum results useless. They are exactly right for a thrown ball over short distances, close enough for a shot put or a long jump, and — crucially — they are the correct starting point that the drag terms are added to. But when you next see a physics answer of "45° for maximum range" applied to a golf ball or an artillery shell, you now know why the real answer is different.
Where this shows up in your life
A basketball player shooting from the free-throw line is solving the trajectory equation, badly and unconsciously, but well enough. The reason a high arc is preferred is not range — it is that a steeper descent presents a larger effective hoop, since the ball approaching at 45° sees an ellipse and one approaching at 70° sees nearly the full circle.
Aircraft navigation is the river problem with wind instead of current. A plane whose nose points due north in a westerly crosswind travels north-east over the ground, and the angle between where the nose points and where the plane actually goes is called the crab angle. Pilots compute it with exactly the triangle above.
And the independence of horizontal and vertical motion is why a passenger can toss a coin straight up inside a cruising aircraft and catch it, rather than having it slam into the rear bulkhead at 900 km/h. The coin keeps the horizontal velocity it had when it left the hand, because nothing horizontal acts on it. Chapter 1.4 makes that into a law.
What the next chapter fixes
We can now describe motion completely — where something is, how fast, in what direction, along any path. What we still cannot do is say why anything moves the way it does. Nothing so far explains why the projectile accelerates downwards at g rather than at some other rate, or why the horizontal velocity stays fixed. Chapter 1.4 introduces the three laws Newton published in 1687, which turn the description into an explanation and make the rest of this Part possible.