Week 3 In-Depth — Why 2D Motion Splits Into Two 1D Problems

Read the cheatsheet first. This note answers “why does any of this work?” and walks through one full worked example per major topic, with every step on the page.

1 — Why orthogonal-plane decomposition works

Newton’s second law in 2D is a vector equation:

$\vec{F}=m\vec{a}.$

A vector equation in 2D is really two scalar equations, one for each component:

$F_x=m{a}_x,F_y=m{a}_y.$

These two equations don’t talk to each other. Whatever force acts in $x$ produces an acceleration in $x$; whatever force acts in $y$ produces an acceleration in $y$. The $x$-motion has no way of knowing about the $y$-motion except via time, which is the same clock for both.

So once you have decided on a coordinate frame, you can treat the $x$-motion and the $y$-motion as two completely separate 1D problems, both running on the same stopwatch. That’s it — that’s the whole insight that makes 2D kinematics tractable.

Why this matters for projectiles

For a projectile (no air resistance) the only force is gravity, which points purely in $-y$ (if you chose $+y$ up):

$F_x=0\Rightarrow a_x=0,F_y=-mg\Rightarrow a_y=-g.$

Horizontal velocity never changes. Vertical velocity falls at $9.81$ m/s every second. That’s why the cheatsheet says $a_x=0$ and $a_y=\pm g$ — it’s not a rule you memorize, it’s what Newton’s law gives you when the only force is vertical.

Why the SUVAT equations exist at all

The four “constant-acceleration” equations are not separate facts. They are four ways to project the same calculus into algebra, when $a$ is constant. Start from

$a=\frac{dv}{dt}\text{(constant)}⟹v(t)=u+at.$

That’s equation 1. Integrate again:

$v=\frac{dx}{dt}⟹s(t)=ut+\frac{1}{2}a{t}^2.$

That’s equation 3. The other two ($v^2=u^2+2as$ and $s=\frac{u+v}{2}t$) are derived by eliminating $t$ or $a$ algebraically from the first two. Each version is convenient when a different variable is “missing.”

This is why SUVAT only works when $a$ is constant — the integration above assumed it. If $a$ varies with time you have to integrate $a(t)$ directly.

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2 — Why the ${\tan }^{-1}$ quadrant correction matters

Your calculator’s ${\tan }^{-1}$ button returns a value in $(-90°,90°)$. But a vector in the plane can point in any of four quadrants. Two different vectors,

$\vec{A}=(3,4)\text{and}\vec{B}=(-3,-4),$

have the same ratio $A_y/A_x=4/3$ — calculator gives ${\tan }^{-1}(4/3)\approx 53°$ for both. But $\vec{A}$ points up-right (quadrant 1) and $\vec{B}$ points down-left (quadrant 3). They are $180°$ apart.

So the rule is:

• Quadrant 1 ($A_x>0$, $A_y>0$): ${\tan }^{-1}$ gives the right answer directly.
• Quadrant 4 ($A_x>0$, $A_y<0$): ${\tan }^{-1}$ gives a negative angle, also correct.
• Quadrants 2 and 3 ($A_x<0$): add $180°$ to fix the flip.

The simplest sanity check: sketch the vector first. If your answer doesn’t roughly point that way, you missed a quadrant correction.

Worked example with all the steps — Example 1, the rifle shot

A rifle is aimed horizontally at a target $50$ m away. The bullet hits $2$ cm below the target. Find muzzle speed and impact velocity.

Setup. Choose $+x$ to the right (toward target), $+y$ downward so gravity is positive: $a_y=+9.8{\text{m/s}}^2$.

	$x$-plane	$y$-plane
$a$	$0$	$+9.8$ m/s${}^2$
$s$	$50$ m	$0.02$ m
$u$	$u_x$ (unknown)	$0$ (horizontal aim)

(a) Flight time from the $y$-plane:

$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2⟹0.02=\frac{1}{2}(9.8)t^2⟹t=\sqrt{\frac{0.04}{9.8}}\approx 0.064\text{s}.$

(b) Muzzle speed. With $a_x=0$, $s_x=u_{x}t$:

$u_x=\frac{50}{0.064}\approx 781.25\text{m/s}.$

(c) Impact velocity. Horizontal is unchanged: $v_x=781.25$ m/s. Vertical from $v_y=u_y+a_{y}t$:

$v_y=0+9.8(0.064)\approx 0.627\text{m/s}.$

Magnitude and direction:

$|\vec{v}|=\sqrt{781.25^2+0.627^2}\approx 781.25\text{m/s},\theta ={\tan }^{-1}\left(\frac{0.627}{781.25}\right)\approx 0.046°$

below the horizontal.

Why this answer makes sense. The bullet barely drops (2 cm in 50 m), so the trajectory is almost flat — and indeed the impact angle is a tiny $0.046°$ off horizontal. The Assess step passes.

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3 — Why the relative velocity rule has the form it does

Imagine you are standing in a river. You swim due north at $4$ m/s relative to the water. The water itself is moving east at $3$ m/s relative to the ground.

Your ground velocity is not “north at 4” and it is not “east at 3.” It is both at once, because in one second:

• You drift $3$ m east (with the water), plus
• You swim $4$ m north (through the water).

So your displacement after $1$ s is $(3,4)$ m, and your ground velocity is $(3,4)$ m/s with magnitude $\sqrt{9+16}=5$ m/s.

In symbols:

${\vec{v}}_{\text{you, ground}}={\vec{v}}_{\text{you, water}}+{\vec{v}}_{\text{water, ground}}.$

Which is exactly the subscript convention ${\vec{v}}_{OG}={\vec{v}}_{OM}+{\vec{v}}_{MG}$, with O = swimmer, M = water (the medium), G = ground. The inner subscripts cancel: $\text{O}\underline{\text{M}}+\underline{\text{M}}\text{G}=\text{OG}$.

Why it splits per plane: because vector addition is component-wise. The $x$-components add separately, the $y$-components add separately.

Worked example — Example 3, the plane in a crosswind

An observer on the ground sees a plane moving due north at $220$ km/h. The wind blows west at $40$ km/h. Find the plane’s velocity relative to the air.

Setup. $+y$ = north, $+x$ = east. Define: O = plane, M = air, G = ground.

Given:

${\vec{v}}_{OG}=0\hat{i}+220\hat{j}\text{km/h},{\vec{v}}_{MG}=-40\hat{i}+0\hat{j}\text{km/h}.$

(Wind blowing “west” means the air’s velocity-vector points in $-x$.)

We want ${\vec{v}}_{OM}$, the plane’s velocity relative to the air (i.e. its heading, the way the pilot is actually pointing the nose). Rearrange ${\vec{v}}_{OG}={\vec{v}}_{OM}+{\vec{v}}_{MG}$:

${\vec{v}}_{OM}={\vec{v}}_{OG}-{\vec{v}}_{MG}.$

Per plane:

$v_{OM,x}=0-(-40)=+40,v_{OM,y}=220-0=+220.$

So ${\vec{v}}_{OM}=40\hat{i}+220\hat{j}$ km/h.

Magnitude and direction.

$|{\vec{v}}_{OM}|=\sqrt{40^2+220^2}\approx 224\text{km/h},\theta ={\tan }^{-1}\left(\frac{40}{220}\right)\approx 10.3°\text{east of north}.$

Intuition check. The wind is pushing the plane west; to still appear to be tracking due north, the pilot must point the nose slightly east of north to cancel that drift. The $+40\hat{i}$ component says exactly that. Good.

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4 — A second projectile, with non-zero launch angle

Worked example — Example 2, the cliff projectile

Initial speed $|u|=185$ m/s at $71°$ above the horizontal; flight time $40.0$ s. No air resistance. Find: (a) speed at $t=21$ s, (b) velocity just before impact, (c) cliff height.

Setup. $+x$ horizontal (direction of launch), $+y$ up. $a_y=-9.8$ m/s${}^2$, $a_x=0$.

Decompose the launch velocity:

$u_x=185\cos 71°\approx 60\text{m/s},u_y=185\sin 71°\approx 175\text{m/s}.$

(a) Speed at $t=21$ s. $v_x$ stays at $60$ m/s forever. Vertical:

$v_y=u_y+a_{y}t=175+(-9.8)(21)=-30.8\text{m/s}.$

Speed:

$|\vec{v}|=\sqrt{60^2+(-30.8{)}^2}\approx 67.4\text{m/s}.$

The negative $v_y$ means the projectile is already past the peak and falling.

(b) Velocity just before impact ($t=40$ s).

$v_x=60,v_y=175+(-9.8)(40)=-217\text{m/s}.$ $|\vec{v}|=\sqrt{60^2+217^2}\approx 225\text{m/s},\theta ={\tan }^{-1}\left(\frac{217}{60}\right)\approx 74.5°\text{below horizontal}.$

(c) Cliff height. Displacement in $y$ at $t=40$:

$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2=175(40)+\frac{1}{2}(-9.8)(40{)}^2=7000-7840=-840\text{m}.$

Negative because we chose $+y$ up — the projectile ends up $840$ m below launch. So the cliff is $840$ m tall.

Why this answer makes sense. At launch the velocity is $185$ m/s at $71°$ — steep. By $t=21$ s it has slowed to $67$ m/s near the top of its arc. By $t=40$ s it has accelerated back to $225$ m/s, almost straight down ($74.5°$ below horizontal). The cliff height of $840$ m is consistent with a $40$-second flight under gravity.

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5 — How these ideas connect

Every concept this week is one consequence of two ideas:

Idea	What it gives you
Vectors decompose into components.	Polar ↔ Cartesian conversions, $\hat{i},\hat{j}$ notation, and the very ability to “split into $x$ and $y$.”
Newton’s law works component-wise.	$x$ and $y$ motions are independent (only $t$ is shared). SUVAT applies in each plane. Projectile motion is just the case where $a_x=0$ and $a_y=\pm g$. Relative velocity is just vector addition of those component velocities.

Master those two ideas and you can re-derive everything else on the cheatsheet from scratch.

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6 — Exam-style sample, end-to-end

A stunt car (Tutorial 3, Exercise 2) drives off the edge of a horizontal cliff at $72$ km/h. It lands $15$ m below. Find: (a) flight time, (b) horizontal distance travelled.

Setup.

Item	Model
Coordinates	$+x$ in direction of motion, $+y$ down
Velocity unit conversion	$72\text{ km/h}÷3.6=20$ m/s
Known $x$-plane	$u_x=20$ m/s, $a_x=0$, $s_x=?$
Known $y$-plane	$u_y=0$, $a_y=+9.8$ m/s${}^2$, $s_y=15$ m
Shared	flight time $t$, identical in both planes

(a) Flight time — from the $y$-plane, since we know everything there:

$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2⟹15=0+\frac{1}{2}(9.8)t^2$

$t=\sqrt{\frac{30}{9.8}}\approx 1.75\text{s}.$

(b) Horizontal distance — substitute that time into the $x$-plane:

$s_x=u_{x}t+\frac{1}{2}{a}_x{t}^2=(20)(1.75)+0=35\text{m}.$

Assess.

• Units: m/s × s = m. Good.
• Sign: positive, as expected (the car moves forward).
• Magnitude: $35$ m forward for a $15$ m drop at $20$ m/s is plausible — the cliff is about half the horizontal distance, which makes sense for a fast launch and a short drop.

Answer. The car lands $\boxed{35\text{m}}$ horizontally from the cliff edge after a $1.75$ s flight.

That layout — Coordinates → Tables per plane → Pick the plane you know completely → Solve for $t$ → Plug $t$ into the other plane → Assess — is the template for every projectile question on the paper.

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Where to look next

• The cheatsheet for fast revision + the quiz.
• The full lecture summary for every formula and worked example as covered in lecture.
• Tutorial 3 + solutions PDF for four more end-to-end problems.
• Wolfson Ch. 2 for textbook reading.