— Physics · AS Level (7408) · Linear Motion & Projectiles —
★ Exam Tip: This topic is tested almost every paper — graph MCQs
in Paper 1, calculation questions with the kinematics equations in Paper 1 & 2, and projectile motion combines
both. Definitions are guaranteed 1-mark bookwork — memorise them exactly (in your own words, keeping the
key phrases), and know by heart what the gradient and area of every s–t, v–t and a–t graph
represent.
1. Displacement & Distance
Definition — Displacement: the shortest distance between two
points, measured from a fixed frame of reference, along a specific linear direction.
Vector, measured in metres.
Remember: to define displacement you always need TWO things — (1) a
fixed frame of reference (the point you measure from — usually the start of the journey), and (2) a
direction (the line of action).
Definition — Distance: the actual length covered during a
journey, from a fixed frame of reference. Scalar, measured in metres — no direction needed.
displacement = straight line A→B; distance = the actual curved path
Classic pairing to remember: a runner completes one full lap of a 200 m track and returns to the start ⇒
distance = 200 m but displacement = 0 m (back at the fixed frame
of reference).
① Resolving displacement along a line of action
When a point R sits off the measuring line, resolve its position vector into components using the
angle-opposite = sin, angle-adjacent = cos rule (head-to-tail), then read off which side of the fixed frame
of reference F you are on along that line.
R on + side, on the line: simple +2resolve: opposite θ → sin, adjacent θ → cos
2. Velocity & Speed
Definition — Velocity: the rate of change of displacement along
a specific linear direction. Vector, measured in m s⁻¹.
Definition — Speed: the rate of change of distance. Scalar,
measured in m s⁻¹ — no direction needed.
v = dsdt = ΔsΔt
Gradient of an s–t graph = velocity.Area under a v–t graph = Δs (since v×Δt = Δs).
gradient of s–t = v; tangent needed where curvedarea under v–t graph = Δs (change in displacement)
Average speed / average velocity = total distance (or
displacement)total time
Sign-convention worked mini-examples (down = +, circled convention):
Motion
Displacement
Velocity
thrown down 2 m/s, returns to start
0 m
+2 m/s
thrown up 2 m/s, returns to start
0 m
−2 m/s
moves down 2 m at 2 m/s
−2 m (above F, using up=+ locally) / +2 m (down=+)
+2 m/s
6 m below F, moving up at 4 m/s
+6 m
−4 m/s
Common mistake: the sign of displacement and the sign of velocity do
NOT have to match — displacement depends where you are relative to F; velocity depends which way you're currently
moving.
3. Displacement–Time Graphs
Method: read the graph "knowing nothing" first — is s positive/negative,
increasing/decreasing? Then find gradient behaviour to get velocity. Don't jump straight to memorised shape
labels.
curving to a plateau: +v decreasing to 0flat then curving up: +v increasing to constantcurving down to 0: −v increasing (leftwards, speeding up)
steep then flattening (still +s): −v decreasing to 0
Common mistake: if s is decreasing but still stays positive (hasn't
crossed the axis), it's still "positive displacement, decreasing" — don't call the displacement itself negative;
it's the velocity carrying it back that is negative.
4. Acceleration
Definition: the rate of change of velocity along a specific
linear direction. Vector, measured in m s⁻². (No fixed frame of reference needed — only a direction, same
as velocity.)
★ Exam Tip: Definitions are guaranteed 1-mark bookwork — memorise them
exactly, but write them in your own understanding rather than parroting the mark scheme verbatim; examiners look
for the key phrase ("...along a specific linear direction"), not an exact word match. Adding "it is a
vector/scalar measured in ___" rarely earns marks itself but costs nothing if you have space/time.
a = dvdt = ΔvΔt
Gradient of a v–t graph = acceleration.Area under an a–t graph = Δv (a×Δt = Δv).
gradient of v–t = aarea under a–t graph = Δv
AS Physics: take g = 9.81 m s⁻² "downwards" (state it as a
magnitude+direction; whether it's + or − depends on YOUR chosen convention) — the acceleration of all objects
experiencing only their weight.
+v decreasing to 0, −a constantv→0, a leftwards & increasingv increasing, a rightwards & increasing
v increasing, a rightwards but decreasingv decreasing to const, a leftwards decreasing to 0
Note: in every case, gradient of v–t = a; a curve flattening (m
decreasing in magnitude) means the acceleration itself is shrinking, even while velocity keeps changing in the
same direction.
6. "Which Graph" MCQ Technique
Trick — Key Pointers: for any "which graph shows…" question, first
mark every point on the GIVEN graph where the gradient = 0 → these become compulsory zero-value points on the
answer graph. Eliminate any option with an extra or missing zero point before even checking overall shape.
Moon featherMCQ · speed–time
Free-fall acceleration on the Moon = 1.6 m s⁻² (no atmosphere). Astronaut drops a feather. Which graph shows
speed vs time during the first second?
Constant a ⇒ speed increases at a constant rate ⇒ straight line through the origin,
gradient = 1.6.
A ✓ straight line, gradient const = 1.6B ✗ curve concave up (a increasing)C ✗ curve concave down (a decreasing)D
✗ horizontal — a=0, wrong
Answer: A
v–t hump → s–t shapeMCQ · displacement from v–t
v–t graph: a hump above the axis then a trough below, both regions bounded by m=0 points (+Δs then −Δs). Which
s–t graph matches, t=0 to T?
Key pointers: m=0 on v–t at start, at the v-axis crossing, and at t=T ⇒ s–t must have zero
gradient (turning points) at exactly these three t-values. s rises while v>0, then falls while v<0.
A ✓ s rises to a peak then falls back to 0 — turning points
matchB ✗ sine wave — extra zero-gradient pointsC ✗ crosses zero at T/2 — v never shown negative thereD ✗ two humps, both v≥0 — ignores the negative region
Answer: A
s16 qp13 q6MCQ · velocity–time
A ball rolls in a straight line up a ramp then back down along the same path. Which graph shows v vs t?
Take up-the-ramp as +. Constant deceleration (gravity component) the whole time ⇒ ONE constant
negative gradient throughout, v starts +, passes through 0 at the top, continues to −v coming back down.
A ✓ one straight line, + → − through 0B ✗ rises then symmetric fall back to 0 — never negativeC ✗ sudden jump — impossible, a is constantD
✗ starts negative → positive — only valid under opposite sign convention
Answer: A (a single constant-gradient line through zero)
w16 qp11 q9MCQ · force→speed
Car stationary at lights; driver presses accelerator sharply. Resultant force jumps from 0 to a constant value.
Which graph shows speed vs time?
Trick: constant mass ⇒ F ∝ a, so a force–time graph has the SAME
SHAPE as an acceleration–time graph (values differ, shape doesn't) — treat it as one.
a = 0 then jumps to a constant value ⇒ v stays 0 momentarily then rises with a constant (nonzero)
gradient — i.e. speed–time must show a corner: flat then straight-line rising.
A ✗ straight line from t=0 — no initial flat
partB ✗ smooth curve — implies gradually changing aC ✓ flat, then rises linearly, gradient constD
✗ subtly wrong initial curvature
Answer: C
w16 qp12 q7MCQ · distance–time
A lorry travels at constant speed then decelerates until it stops. Which graph shows distance vs time?
Constant speed ⇒ first part must be a straight line (constant gradient). Decelerating to a stop ⇒
gradient falls to 0 at the end.
Trick: distance can never curve back down towards zero — it can
only keep increasing (or flatten off), since you keep adding to the total path length. This instantly rules out
any option where the curve returns to 0 (that shape would only fit a displacement–time graph of an object
returning to start).
A ✗ starts curved, not a straight lineB ✓ straight line then flattens to m=0C ✗ wrong shape (falls) — invalid for distanceD ✗ falls back to 0 — invalid for distance
Answer: B
s17 qp13 q7MCQ · force→speed
Driver stops the car in time t by gradually increasing the braking force (F–t graph: straight line
rising from 0). Which graph shows speed vs t?
F∝a (const mass) ⇒ deceleration itself increases with time ⇒ speed–time gradient must get steeper
(more negative) as t increases — a curve concave-down, falling faster and faster to 0.
A ✗ gradient magnitude decreasing — wrong
trendB ✗ straight line — implies constant a, but a is
increasingC ✓ gradient magnitude increasing, curve flattens into the
axisD ✗ initial gradient too shallow / wrong curvature
Answer: C
w17 qp12 q7MCQ · same-displacement trains
Train's v–t graph given (rise, plateau, fall). Another train has double the acceleration, same max
speed, same deceleration. Which v–t graph (same scale) fits the SAME two stations?
Trick: "same two stations" ⇒ same total displacement ⇒ same
area under the v–t graph. Compute the original area with the trapezium formula, Area = ½(a+b)×height,
then check which option reproduces that same area.
Original: A=5, B=11, h=8 → Area = ½(5+11)(8) = 64 unit²Option A → recompute → 56 unit² ✗Option B → recompute → ½(6+10)(8) = 64
unit² ✓
Answer: B (equal area under the curve = equal displacement)
m19 qp12 q6MCQ · spring bounce
Mass on a spring bounces up/down from release at t=0; distance–time graph given (distance up = +). Which graph
shows velocity vs time?
Mark every t where the given graph has m=0 → these are the compulsory v=0 points. Then check
sign: velocity is + while distance-up is increasing, 0 at each peak/trough of distance, and − while returning —
watch for options showing extra/missing zero-crossings.
A ✓ correct zero-crossings and signs match at every marked
pointB/C/D ✗ each has an extra or misplaced zero-velocity point not
seen on the original graph
Answer: A
s22 qp11 q5MCQ · average vs instantaneous acceleration
Curved line PQR is the v–t graph for a car from rest. What is the average acceleration over the first 5
s (P to Q, where Q is at t=5s)?
Common mistake: "instantaneous acceleration at 5 s" (tangent at Q) ≠
"average acceleration over the first 5 s" (straight chord P→Q). Read the wording carefully!
A ✗ area below curve PQ — gives displacement, not
accelerationB ✗ area of triangle PQS — also a displacement-type
quantityC ✓ gradient of the straight (chord) line PQD
✗ gradient of tangent at Q — that's the instantaneous value
Answer: C
s22 qp11 q22MCQ · KE from displacement–time
Sound wave: displacement–time graph of an air particle (period T). Which graph shows kinetic energy vs time?
Trick: KE = ½mv², and v = gradient of s–t. Mark the s–t
zero-gradient points (⇒ v=0 ⇒ KE=0 there). Since KE ∝ v², it is always ≥ 0 — the KE–t graph stays entirely
on/above the axis, and it oscillates at TWICE the frequency of v (KE=0 at every v=0, both up- and
down-crossings).
A/B/C ✗ miss a required zero-KE point, or don't double the
frequencyD ✓ all zero points match, always ≥0, double-frequency
humps
Answer: D
7. Deriving the Kinematics Equations
★ Exam Tip: You need to know the derivation of each equation —
this can appear directly as a Paper 2 question. All three are derived from ONE v–t graph.
the master v–t graph: (0,u) to (t,v), straight line since a is constant
① From the gradient (a = gradient of v–t)a = v
− ut ⇒ v = u + at
② From the area (s = area under v–t, a trapezium)s = ½(u + v)t, sub. v
= u+at ⇒ s = ut + ½at²
③ Eliminating time (combine ① and ②)t = v−ua substituted into ② and simplified ⇒ v² = u² +
2as
Where: u = initial velocity | v = final velocity | a = acceleration | s
= displacement | t = time
Common mistake: deriving ③ needs patience — expand (v−u)² carefully
term by term; rushing this algebra is the most common place to go wrong.
How to solve Kinematics Questions:
Always make a sign convention (before writing any equation!)
Select the best equation (match knowns/unknowns)
Put signs in correctly (substitute known directions; let the algebra return signs of unknowns)
Trick: if a single equation has 2 unknowns, you need a 2nd equation from
another stage of the same motion — solve simultaneously, just like ordinary algebra.
Trick: only assign a sign yourself to a quantity whose direction you
already know. For anything you're solving FOR, let the kinematics equation's algebra return the correct sign
automatically — don't guess it in advance.
8. Worked Problems — Using the Equations
w17 qp12 q6Calculation · vertical motion
Hot-air balloon moves up at constant 3.00 m/s. A sandbag is dropped; it takes 5.00 s to fall to the ground.
What was the height of the balloon when dropped?
Convention: down = +. Sandbag shares the balloon's velocity at release, i.e. u = −3 (upwards).
s = ut + ½at²h = (−3)(5) + ½(9.81)(5)² = −15 +
122.6 = 107.6 ≈ 108 m
Answer: B (108 m)
Goods vs express trainCalculation · overtaking
Goods train passes a station at steady 10 m/s. Express train (at rest there) starts with uniform acceleration
0.5 m/s² just as the goods train passes. How long until the express overtakes it?
Both trains cover the SAME displacement s when they meet.
Goods: s = ut = 10tExpress: s = ut + ½at² =
½(0.5)t² = 0.25t²Equate: 10t = 0.25t² ⇒ t(0.25t − 10) = 0 ⇒ t = 40 s
Answer: D (40 s)
Acceleration between X, Y, ZCalculation · simultaneous
equations
Uniform acceleration in a straight line. XY = YZ = 40 m. Time X→Y = 12 s, Y→Z = 6.0 s. Find the acceleration.
Trick: two unknowns (u and a) in one equation ⇒ need a 2nd
equation. Take the journey X→Z as a whole (not Y→Z) so the SAME initial velocity u (at X) appears in both
equations.
Trains arrive at 50 m/s. At marker 1, driver brakes uniformly to pass marker 2 at ≤10 m/s. Emergency brake
triggers if the time between markers is <20 s. How far apart should the markers be?
Convention: forward = +. Two unknowns (a, s) ⇒ two equations.
① v=u+at: 10 = 50 + a(20) ⇒ a = −2 m s⁻²② s=ut+½at²: s = 50(20) + ½(−2)(20)² = 1000 − 400 = 600 m
Trick: the algebra itself returned a negative a — confirming the
deceleration direction — without you having to guess the sign in advance.
Train always passes a yellow light (speed u) before braking uniformly to stop exactly at a red light, distance
x away. If speed is increased by 20% (deceleration unchanged), what is the new minimum distance?
★ Major Exam Tip — "Comparing questions": ~3–4 Paper 1 questions
are this type: describe scenario 1, extract a relation, then tweak ONE variable and ask for the new answer in
terms of the original. Method: (1) solve case 1 symbolically, isolate the constant piece; (2) redo for
case 2 with the changed variable; (3) spot the case-1 expression hiding inside case 2's result and substitute it
back in.
Case 1: v²=u²+2as ⇒ 0 = u² + 2(−a)x ⇒ x = u²/2aCase 2 (u→1.2u): 0 = (1.2u)² + 2(−a)x₂ ⇒ x₂ = 1.44u²/2aBut u²/2a = x (from case 1) ⇒ x₂ = 1.44x
Answer: C (1.44x)
s21 qp11 q7Calculation · multi-stage v–t graph
Train: rest → uniform a=0.20 m/s² up to 20 m/s → constant speed → uniform decel 0.40 m/s² to rest. Stations
3000 m apart. Total time?
Trick: for multi-stage motion (accelerate + cruise + decelerate),
draw the v–t graph FIRST — turns three separate calculations into one shape you read areas/gradients off.
t₁ (accel phase): v=u+at ⇒ 20 = 0.2t₁ ⇒ t₁ = 100 sdecel phase duration: 0 = 20 + (−0.4)t ⇒ t = 50 sArea (trapezium, total time T, parallel sides T and (T−150), height 20): 3000 = ½[T +
(T−150)]×206000 = (2T−150)(20)/20 → 2T − 150 = 300 → T = 225 s
Note: the exact working gives 225 s; the closest official MCQ
option is 230 s (C) — treated as rounding in the given options, not an error in the method.
Answer: C (230 s, nearest to the calculated 225 s)
Polystyrene sphere falls from rest; flash photos every 0.1 s. (i) How can you tell it reaches constant speed?
(ii) Distance fallen at 0.7 s and 1.1 s?
(i) The distance between consecutive dots becomes equal.
(ii)-1: count dots to 0.7 s from the scale → 132 cm(ii)-2: at constant speed, each 0.1 s afterwards adds 25 cm → 132 + 4(25) = 232 cm
Answers: 132 cm; 232 cm
Lead sphere — no terminal velocityCalculation · counting
photos
Same setup, lead sphere (constant acceleration, no terminal speed reached). How many flash photos appear
against a 160 cm scale (include the t=0 photo)?
s = ut + ½at² (u=0): 160×10⁻² = ½(9.81)t² ⇒ t = 0.571 sn = 0.571/0.1 = 5.7 ⇒ 5 intervals + the t=0 photo = 6
Answer: 6 photographs
Toy car on a slope (d vs t²)Calculation · graph gradient → a
d (y-axis) plotted against t² (x-axis) for a toy car released from rest on a slope. (a) Random error → scatter
of points. Systematic error → non-zero t² intercept (false origin). (b) Find acceleration from the gradient.
s = ut + ½at² (u=0) ⇒ s = ½at², i.e. y = (½a)x — gradient = ½agradient = (100−20)/(10−2.4) = 10.5 cm s⁻²½a = 10.5×10⁻² ⇒ a = 0.212 m s⁻²
Answer: gradient 10.5 cm s⁻²; a = 0.212 m s⁻²
s22 qp22 q3 — ball thrown from a wallCalculation · vertical
motion
Man on a wall throws a ball up at 5.6 m/s from a height of 3.1 m. (a) Show max height above ground = 4.7 m. (b)
Time to fall from max height to ground. (e) Heavier 2nd ball, same u, same h — compare acceleration and landing
speed.
(a) v²=u²+2as: 0=(5.6)²+2(−9.81)s ⇒ s=1.598 m; 1.598+3.1 ≈ 4.7 m ✓(b) s=ut+½at²: −4.7 = ½(−9.81)t² ⇒ t = 0.98 s
(e)(i) Same acceleration (g is independent of mass). (e)(ii) Same speed hitting the ground
(energy/kinematics don't depend on mass here).
Girl & boy on bicyclesCalculation · catch-up, symbolic
Girl G rides at constant 3.5 m/s, passing stationary boy B at t=0. B accelerates uniformly to 5.6 m/s in 5.0 s,
then holds 5.6 m/s. At t=T he catches her. Find T.
Braking car — half initial speedCalculation · proportion (two
stages)
Car brakes in two stages: stage 1 constant 18 m/s (0.65 s), stage 2 uniform deceleration 18→0 (2.85 s). (b)
total distance 0→3.5s. (c) if the SAME car braked from half the initial speed (same deceleration), how does
distance change in each stage?
(b) Area = rectangle+triangle = (18×0.65) + ½(2.85)(18) = 37.35 ≈ 37.4
m
Stage 1 (constant v): s = vt (t fixed) ⇒ s ∝ v ⇒ halving v halves
s.
Stage 2 (uniform decel to 0): v²=u²+2as with a constant ⇒ u² ∝ s.
Halving u: (½u)²/s₂ = u²/s ⇒ s₂ = ¼s — distance becomes a QUARTER, not a half.
Light-gate ball & diameterCalculation · common mistake
Steel ball (diameter 0.080 m) released from rest, falls 0.280 m to a light beam. (ii) Time to reach the beam.
(iii) Time interval during which the beam stays broken.
Common mistake: for (iii), do NOT restart the calculation with u=0
at the point the beam first breaks — u is only 0 at the very start of the fall. Instead find the time for the
FULL distance (to beam + one diameter) from the true release point, then subtract the shorter time.
(ii) s=ut+½at² (u=0): 0.280 = ½(9.81)t² ⇒ t = 0.239 s(iii) full distance 0.080+0.280=0.360: 0.360=½(9.81)t² ⇒ t=0.271 sinterval = 0.271 − 0.239 = 0.032 s
Answers: 0.239 s; 0.032 s
Measuring g with light gatesMethod · derivation
Electromagnet releases an iron ball in a vacuum (removes air resistance). It falls distance x to light-gate 1
(time t₁), then a further h to light-gate 2 (time t₂). Derive an expression for g.
① x = ½g t₁²② x + h = ½g t₂²② − ①: h = ½g(t₂² − t₁²) ⇒ g = 2h/(t₂² − t₁²)
g = 2h / (t₂² − t₁²) → matches MCQ answer D for the equivalent "two markers M₁, M₂" question
9. Projectile Motion
★ Exam Tip: ALWAYS state a convention for BOTH axes before
touching any equation — e.g. "upwards is positive AND rightwards is positive." Skipping this is the #1 source of
sign mistakes in this topic.
Trick: pick your convention to match the ball's actual initial
direction of travel — this minimises negative signs in your working (fewer chances to slip). The final numeric
answer is identical whichever convention you pick — "maths never fails, only bookkeeping does" — but fewer minuses
means fewer mistakes.
v(x) constant throughout (no horizontal force); v(y) decreases, 0 at top, then reverses
Horizontal: a = 0 always (weight has no horizontal component, air resistance
neglected) ⇒ vx is constant ⇒ x = vxt.
Vertical: a = g always — never zero, not even at the top of the path (common
wrong-option trap). At max height, vy=0 but vx≠0 and a≠0.
x = (v cosθ) t | y = (v sinθ) t − ½gt² |
Hmax = (v sinθ)²2g
Symmetry trick: if a projectile lands at the SAME height it launched
from, vy at a given height going up = −(vy at that height coming down). Resultant speed is
the same magnitude, only the angle differs. This breaks if launch/landing heights differ.
Toy car ramp-to-rampCalculation · resolving components
Toy car leaves ramp P at 5.5 m/s at angle θ; horizontal component 4.6 m/s; lands at the top of ramp Q (same
height), distance d apart.
(i) Pythagoras: 5.5² = 4.6² + v_y² ⇒ v_y = 3.0 m/s (alt: v_y=5.5sinθ once θ
found from cosθ=4.6/5.5)(ii) s_y=v_yt+½at²: 0 = 3t + ½(−9.81)t² ⇒ t(4.905t−3)=0 ⇒ t=0.6116 ≈ 0.61 s(iii) d = v_x t = 4.6(0.6116) = 2.8 m
v_y=3.0 m/s; t=0.61 s; d=2.8 m
w18 qp22 q1 — golferCalculation · components
Ball leaves ground at 6.0 m/s at angle θ; vertical component 4.8 m/s. (a) Show horizontal component = 3.6 m/s.
(c) Max height.
m22 qp22 q2 — hosepipe dropletCalculation · full projectile
Water leaves a hosepipe at P horizontally at 6.6 m/s, height h above ground, lands at Q, 3.5 m away. (a) Why is
v_x constant? (b) Show t=0.53s. (c) Find h. (e) Find displacement P→Q.
(a) Air resistance negligible and no other horizontal force ⇒ a=0 ⇒ v constant.
Note (d): displacement P→Q is SMALLER than the distance
travelled — displacement is the straight line P to Q, distance is the actual curved path.
t=0.53s; h=1.4 m; displacement=3.8 m
m16 qp12 q8 — ball across a fieldMCQ · statement
Which statement describes the motion (air resistance ignored)?
A ✗ lands with same velocity as thrown — only true for equal
launch/land height AND same speed, but here direction differs too broadly statedB ✓ horizontal component of acceleration constant (=0) throughoutC ✗ "vertical AND horizontal accel both zero at highest point" — vertical accel
is NEVER zero (always g)D ✗ horizontal AND vertical velocity both zero at top — only v_y=0,
v_x≠0
Answer: B
w16 qp11 q7 — astronaut on the MoonCalculation · resultant
velocity
Ball thrown with v_y=8.00 m/s, v_x=4.00 m/s; Moon g=1.62 m/s². Speed 9.00 s after being thrown?
s16 qp13 q7 — displacement formulas tableMCQ · formula table
Ball thrown with velocity V at angle θ. What are horizontal displacement x and vertical displacement y after
time t?
Common mistake: never attempt these without a stated convention
first — mixing up sinθ/cosθ or missing the sign on g is very easy otherwise.
x = (Vcosθ)t (horizontal, a=0)y = (Vsinθ)t −
½gt² (vertical, a=−g if up is +)
Answer: B — x=Vtcosθ, y=Vtsinθ − ½gt²
s18 qp12 q6 — speed–time graph of a projectileMCQ · graph
shape
Which graph best shows the variation with t of the SPEED of a projectile, launch to landing (level ground)?
Speed = √(v_x² + v_y²). v_x is constant and non-zero throughout, so speed can NEVER be constant
(v_y changes) and NEVER reaches zero (v_x alone keeps it above zero, even at v_y=0). The relationship isn't
linear in t, so it must be a smooth curve, not a straight line or a V-shape.
A ✗ constant — wrong, v_y changesB ✗ V-shape touching zero — speed never reaches 0C ✓ smooth curve, dips to a minimum (=v_x) at the top, rises
againD ✗ touches zero at the bottom — impossible
Answer: C
m21 qp12 q7 — cannon max heightCalculation · derivation
Cannon fires at speed v, angle α. Which equation gives max height H?
At max height v_y=0. v²=u²+2as: 0² = (vsinα)² + 2(−g)H⇒ H = (vsinα)²/2g
Answer: C — H = (v sinα)²/2g
✓ Quick Recap — Full Chapter Cheat-Sheet
Graph
Gradient gives
Area under gives
s–t
velocity
—
v–t
acceleration
Δs (displacement)
a–t
—
Δv (change in velocity)
Displacement = shortest distance, fixed frame of reference + direction, vector, m. Distance =
actual path length, scalar, m.
Velocity = rate of change of displacement (vector, m/s). Speed = rate of change of distance
(scalar, m/s).
Acceleration = rate of change of velocity (vector, m/s²) — no fixed frame of reference needed.
g = 9.81 m/s² downwards (AS Physics) — sign depends on YOUR convention.