DSE Physics | Projectile motion - tutor prep guide | Two axes, the time bridge, four common mistakes

DSE Physics | Projectile motion – tutor prep guide | Two axes, the time bridge, four common mistakes

2026-11-05

Tutor lesson-prep guides · Secondary Physics

The moment students see a projectile they hunt for a projectile formula, and no such formula exists. What works is two sets of straight-line equations plus one rule: treat the axes separately and join them with time. The lesson turns that rule into a reflex.

Objectives — Students can: (1) resolve an initial velocity into horizontal and vertical components and state the acceleration on each axis; (2) use the single flight time to link the axes, finding time before distance; (3) in an angled launch, apply the condition at the top (vertical velocity zero) and compute the range.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Resolving vectors	What is the horizontal component of 20 m/s at 30 degrees?	17.3 m/s, using cos 30
Constant-acceleration equations	What does each symbol in v = u + at mean?	Final velocity, initial velocity, acceleration, time
Free fall	Ignoring air resistance, what is the vertical acceleration?	9.81 m/s2, downwards

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-6 min	One warm-up	One ball dropped, one thrown horizontally from the same height: which lands first?	Guess first, then explain
6-20 min	Resolve the axes	Split u into ux and uy, write the acceleration on each axis	Fill a two-axis data table
20-32 min	The time bridge	Get t from the vertical axis, hand it to the horizontal axis	Work the 45 m platform example
32-44 min	Angled launch	Use uy for the top and the total time, then multiply by ux	Work the 30 degree example and check units
44-56 min	Four question types	One worked item each, always writing knowns before choosing an equation	Do one item alone and self-mark
56-60 min	Exit check	Collect three items, mark at once	Work alone

03 Teaching script (read it out)

Scene 1: Act one: resolve before you choose an equation

The first reflex is to look for a projectile formula and, failing that, to combine everything into one line. Narrow the field instead. A projectile has only one convention: horizontally there is no force (ignoring air resistance), so the motion is uniform; vertically gravity alone acts, so the acceleration is 9.81 m/s2 downwards. Split the initial velocity into ux = u cos(theta) and uy = u sin(theta), write them on two lines, and use the straight-line equations on each.
For a horizontal launch uy is zero: write the zero, do not leave the cell blank, because that zero is the starting point for the time. Every question begins with the same two-row table, unknowns marked with a question mark, so nobody mixes the axes inside one equation.

Ask the student to write ux and uy for 20 m/s at 30 degrees, then write the acceleration under each.

Scene 2: Act two: time is the only bridge

Time is the only quantity the two axes share, so the order is fixed: find t from the vertical axis, then hand t to the horizontal axis. From a height h dropped freely, t = sqrt(2h/g). With h = 45 m and g = 9.81, t is about 3.03 s; the same t gives range = ux t = 20 x 3.03, about 60.6 m.
Emphasise that time cannot come from the horizontal axis: horizontal motion is uniform, so there is no second known quantity there. The commonest error is dividing range by speed and calling it time, which is using an unknown to find an unknown. Write both orders side by side so the class sees that only one road exists.

Ask the student to keep a two-column page: vertical finds t on the left, horizontal uses t on the right.

Scene 3: Act three: an angled launch adds one condition – vertical velocity is zero at the top

With an angled launch uy is not zero, so time starts from uy: total flight t = 2uy/g, the top is at t/2, and the condition there is vy = 0, giving height = uy squared over 2g. For u = 20 m/s at 30 degrees, uy = 10.0 m/s, ux = 17.3 m/s, t is about 2.04 s, range about 35.3 m, and the top about 5.10 m.
Require the line vy = 0 before any working. Students tend to treat the top as the place where speed vanishes altogether, so they zero the horizontal component too and the range collapses to nothing. Show that absurd answer and let them hear why it cannot be right.

Ask the student to say which velocity is zero at the top and which is not.

04 Question types and worked steps

Type	Example	Steps	Answer
Horizontal launch	From a platform, find the flight time and range	Vertical for t, then horizontal with t	t = sqrt(2h/g), range = ux t
Angled range	Given u and the angle, find the range	Resolve, t = 2uy/g, multiply by ux	Range = ux x 2uy/g
Top of the flight	Find the maximum height or the time to reach it	Set vy = 0 and use vy squared = uy squared – 2gh	h = uy squared / (2g)
Graph question	Read values from an s-t or v-t graph	Decide which axis the graph shows, then read slope or area	Slope of v-t is acceleration, area is displacement

Sanity check (teach this): Sanity check: one layout only all lesson – vertical on the left, horizontal on the right – and go slowly rather than let the axes share a line; use g = 9.81 everywhere and insist on units; an angled launch adds exactly one condition, so do not teach every variant at once; a wrong answer demonstrated is worth more than a right one, because students must hear the absurdity before they change.

05 Independent practice (3 questions)

Question	Answer	What to watch
Resolve	Split 25 m/s at 40 degrees into ux and uy	Check for sin and cos swapped
Time bridge	From 20 m, thrown horizontally, find t and range	Check that horizontal was not used for time
Top of flight	Find the maximum height of the same projectile	Check that vy = 0 was set

06 Common mistakes and what to say

Mistake	Student writes	Say this
Taking time from the horizontal axis	Dividing range by ux and calling it flight time	Horizontal motion is uniform, so that is an unknown over an unknown. Time comes from the vertical axis, where there is acceleration.
Zeroing every velocity at the top	The speed at the top is zero, so the range is zero	Only the vertical velocity is zero at the top; the horizontal velocity is still ux. If it stopped, how would it get across?
Swapping sin and cos	For 20 m/s at 30 degrees, writing ux = 10 and uy = 17.3	The angle is measured from the horizontal, so the side along the horizontal takes cos. Draw the triangle and mark the adjacent side.
Ignoring sign direction	Writing a downward displacement as positive while using g = +9.81	Choose the positive direction first, then write the sign in every cell. Which way did you choose?

07 Homework (paper, section and question numbers)

Secondary Physics chapter index — All chapters in the subject
Prep guide master index — Similar chapters in other subjects
Redo this chapter — Redo one horizontal launch and one angled launch with the two-column layout

08 Exit check (last 5 minutes)

Question	Answer	Decision
Acceleration on each axis	Horizontal 0, vertical 9.81 downwards	Wrong = back to act one
Which axis gives time	The vertical axis	Answering horizontal = reteach act two
Condition at the top	Vertical velocity zero, horizontal unchanged	Answering all zero = reteach act three

09 Tutor FAQ

Which equations must be memorised?

No projectile-specific formula is needed. Three straight-line equations plus axis resolution will generate everything else.

Should we use 9.8 or 9.81?

Follow the value printed on the paper. This lesson uses 9.81 throughout and requires students to state it.

What about air resistance?

DSE ignores it by default. If a question mentions it, expect conceptual judgement only, not calculation.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.