PHY131H1/Unit 3

Unit 3 · Introduction to Physics I

Motion in Two Dimensions

One idea carries the whole unit: horizontal and vertical motion are independent. A projectile falls exactly as though it had been dropped, while moving sideways exactly as though gravity were absent. Two one-dimensional problems, sharing only a clock.

By the end you can

• Split a two-dimensional problem into independent axes.
• Solve for range, maximum height and flight time.
• Explain why $45$ degrees maximises range on level ground.
• Convert between reference frames moving at constant velocity.

Why it matters & where it comes from

Why it exists

Independence of components is what makes vectors worth the trouble. Once you accept that the axes do not talk to each other, every two-dimensional problem reduces to Unit 2 done twice.

In the real world

Ballistics tables, basketball arcs, water fountain design. Relative motion is why an aircraft must aim upwind of its destination, and why the ground speed shown to passengers differs from the airspeed the pilot flies.

Story & open questions

Galileo argued for independence with a thought experiment: a ball dropped from a ship's mast lands at the mast's foot, not behind it, because it keeps the ship's horizontal motion. The same reasoning answered why we do not feel the Earth moving.

Independence of components

Rule The axes do not interact

For projectile motion with air resistance neglected:

Horizontal: $a_x = 0$, so $x = v_{0x} t$ — constant velocity.

Vertical: $a_y = -g$, so all four constant-acceleration equations from Unit 2 apply unchanged.

The only thing shared is $t$. Solve one axis for time, substitute into the other.

Example The classic demonstration

One ball is dropped from a table; another is launched horizontally from the same height at the same instant.

Both have $v_{0y} = 0$ and $a_y = -g$, so both satisfy the same vertical equation. They land at the same time.

The launched ball also travels sideways, so it lands further away — but not later. Horizontal motion cannot affect vertical fall.

Initial speed is not initial vertical speed

Launching at speed $v_0$ and angle $\theta$ means $v_{0x} = v_0\cos\theta$ and $v_{0y} = v_0\sin\theta$. Feeding $v_0$ into a vertical equation is the most frequent error in this unit. Resolve first.

Projectile motion

Rule The standard results

Launched from ground level at speed $v_0$, angle $\theta$, landing at the same height:

Time of flight $t = \dfrac{2v_0\sin\theta}{g}$

Maximum height $H = \dfrac{v_0^2\sin^2\theta}{2g}$

Range $R = \dfrac{v_0^2\sin 2\theta}{g}$

Derive rather than memorise: set $y = 0$ to find the flight time, set $v_y = 0$ to find the peak, and substitute the flight time into the horizontal equation for range.

Example Why 45 degrees

Range depends on $\sin 2\theta$, which is largest when $2\theta = 90$ degrees, so $\theta = 45$.

Because $\sin 2\theta$ is symmetric about that peak, $30$ and $60$ degrees give the same range — one high and slow, one flat and fast.

This holds only for level ground with no air resistance. Real projectiles peak nearer $35$ degrees, and throwing off a cliff shifts the optimum lower still.

Example Off a cliff

A ball leaves a $45$ m cliff horizontally at $15$ m/s. Where does it land?

Vertical: $-45 = 0 - \tfrac12(9.8)t^2$, so $t^2 = 9.18$ and $t = 3.03$ s.

Horizontal: $x = 15 \times 3.03 = 45$ m from the base of the cliff.

Landing speed: $v_x = 15$ throughout, and $v_y = -9.8 \times 3.03 = -29.7$ m/s, so $|v| = \sqrt{15^2 + 29.7^2} = 33$ m/s. Combine components, never add speeds.

The range formula has conditions

$R = v_0^2\sin 2\theta / g$ assumes launch and landing at the same height. Off a cliff or into a basket it is wrong. When in doubt, go back to the two axes and solve from scratch — it is four lines.

Relative motion

Rule Velocities add as vectors

The velocity of A relative to C is the velocity of A relative to B plus the velocity of B relative to C. Written with subscripts, $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$ — the inner subscripts match and cancel.

Also $\vec{v}_{BA} = -\vec{v}_{AB}$: reversing the frame reverses the velocity.

Example Crossing a river

A boat moves at $4$ m/s relative to water; the current is $3$ m/s downstream. The river is $100$ m wide.

Aimed straight across: the crossing takes $100/4 = 25$ s, and the current carries it $3 \times 25 = 75$ m downstream. The current cannot affect crossing time, because it has no component across the river.

To land directly opposite: aim upstream so the across-component is $\sqrt{16 - 9} = 2.65$ m/s. Crossing now takes $100/2.65 = 37.8$ s — slower, because part of the boat's speed is spent fighting the current.

Example Aircraft in wind

A plane flies at $200$ km/h relative to air; a $50$ km/h wind blows from the west. To travel due north, which way should it point?

The westerly wind pushes east, so the plane must aim west of north by $\arcsin(50/200) = 14.5$ degrees.

Ground speed is then $\sqrt{200^2 - 50^2} = 194$ km/h — less than the airspeed. Any crosswind costs you speed.

Common pitfalls

• Using $v_0$ where $v_0\sin\theta$ belongs. Resolve before substituting.
• Applying the range formula when launch and landing heights differ.
• Thinking a projectile's horizontal velocity changes. Without air resistance it is constant for the entire flight.
• Adding speeds instead of components when finding landing speed.
• Getting relative-velocity subscripts backwards. Check that the inner ones match.

Practice

Work each fully before opening the answer. Original exercises written for this guide in the style of the course's problem sets.

Practice 1 · Same time

A bullet is fired horizontally as another is dropped from the same height. Which lands first?

Answer

Neither — they land together. Vertical motion is unaffected by horizontal velocity.

Practice 2 · Components

A ball is launched at $20$ m/s, $30$ degrees. Find $v_{0x}$ and $v_{0y}$.

Answer

$v_{0x} = 20\cos 30 = 17.3$ m/s, $v_{0y} = 20\sin 30 = 10$ m/s.

Practice 3 · Flight time

For that ball on level ground, how long is it in the air?

Answer

$t = 2(10)/9.8 = 2.04$ s. Only the vertical component matters.

Practice 4 · Equal ranges

Which other angle gives the same range as $25$ degrees?

Answer

$65$ degrees, since $\sin 2\theta$ is symmetric about $45$. The pair always sums to $90$.

Practice 5 · Off a table

A ball rolls off a $1.2$ m table at $3$ m/s. How far from the table does it land?

Solution

Fall time from $1.2 = \tfrac12(9.8)t^2$ is $t = 0.495$ s; horizontal distance $3 \times 0.495 = 1.5$ m.

Practice 6 · Landing speed

For that ball, what speed does it land at?

Solution

$v_x = 3$, $v_y = 9.8 \times 0.495 = 4.85$ m/s, so $|v| = \sqrt{9 + 23.5} = 5.7$ m/s. Faster than it left the table, because it gained vertical speed.

Practice 7 · River crossing

Does a current change how long it takes to cross a river when aiming straight across?

Answer

No. The current is parallel to the bank and has no across-river component, so it only changes where you land.

Practice 8 · Into a window

A ball thrown at $25$ m/s and $50$ degrees must pass through a window $12$ m up. Show it can, and find when.

Solution

$v_{0y} = 19.2$ m/s. Solve $12 = 19.2t - 4.9t^2$: the quadratic gives $t = 0.83$ s and $t = 3.09$ s — once rising, once falling. Two real roots confirm the ball does reach that height; a negative discriminant would have meant it never gets there.

Expanded topic checklist

Course coverage. This unit is part of the complete PHY131H1 topic map. It should cover the following items before you move on:

• Independence. Horizontal and vertical motion share only the clock.
• Projectiles. Flight time, peak height, range, and the conditions under which the range formula holds.
• Optimal angle. Why $45$ degrees, and why complementary angles tie.
• Relative motion. Vector addition of velocities and the subscript convention.

Textbook/source alignment

This completes the calendar's "classical kinematics". Check the syllabus on Quercus for assigned problems; projectile questions appear on essentially every term test.

Core idea map

Resolve into components, apply Unit 2 to each axis separately, link them through time, recombine at the end. That sequence solves every problem in this unit and most of Unit 4.

Study routine. Draw two columns, one per axis, and fill in what you know in each. The shared time is the bridge, and seeing the columns side by side makes obvious which axis to solve first.

Deep practice extensions

Use these after the main practice set. They push on definitions and edge cases rather than routine calculation.

Extension 1 · Horizontal velocity

How does a projectile's horizontal velocity change during flight?

Guide

It does not, with air resistance neglected. There is no horizontal force, so $a_x = 0$.

Extension 2 · Range conditions

When does $R = v_0^2\sin 2\theta / g$ apply?

Guide

Only when launch and landing are at the same height, with no air resistance.

Extension 3 · Complementary angles

Why do $30$ and $60$ degrees give equal ranges?

Guide

Because $\sin 2\theta$ takes the same value at both — $\sin 60$ equals $\sin 120$.

Extension 4 · Frame subscripts

What is $\vec{v}_{AB}$ in terms of $\vec{v}_{BA}$?

Guide

Its negative. Swapping which object is the reference reverses the relative velocity.

← Unit 02Kinematics in one dimension Unit 04 →Newton's laws