Projectile motion combines horizontal uniform motion with vertical motion under gravity. The key insight — fundamental to all of Mechanics 1 — is that horizontal and vertical components are completely independent. You treat each direction separately using SUVAT, then combine results. This topic appears in virtually every Cambridge 9709 M1 paper.
x = Ucosα · t | y = Usinα · t − ½gt² | g = 9.8 m s⁻² | R = U²sin2α / g | H = (Usinα)² / (2g)
Learning Objectives
Resolve initial velocity into horizontal and vertical components: Ucosα and Usinα
Apply equations of motion independently in each direction
Find time of flight, maximum height and range on level ground
Derive and use the trajectory equation y = x tanα − gx²/(2U²cos²α)
Solve projectile problems from a height above the ground
Find the velocity (magnitude and direction) at any point during flight
Determine when a projectile passes through a given point to find U or α
Show that the maximum range on level ground occurs at α = 45°
Independence Principle
Horizontal: no acceleration. Vertical: g = 9.8 m s⁻² downward
A projectile is a particle launched into the air with an initial velocity, subject only to gravity. We ignore air resistance throughout A-Level M1. The particle follows a curved (parabolic) path.
Key assumptions in A-Level Mechanics 1:
• The projectile is a particle (no size, no rotation)
• Air resistance is ignored
• Gravity acts vertically downward at g = 9.8 m s⁻² (constant)
• The ground is flat and horizontal (unless stated otherwise)
Resolving the Initial Velocity
If a particle is launched with speed U at angle α above the horizontal, we split the velocity into two perpendicular components:
Horizontal component: Ux = U cosα Vertical component: Uy = U sinα (upward)
Example: A ball is kicked at 20 m s⁻¹ at 30° above the horizontal.
Horizontal: 20 cos30° = 20 × (√3/2) = 17.32 m s⁻¹
Vertical: 20 sin30° = 20 × 0.5 = 10 m s⁻¹ (upward)
To find components from a diagram: adjacent side → cos, opposite side → sin.
If the angle is measured from the vertical (not horizontal), swap sin and cos.
The Independence Principle
This is the most important concept in projectile motion. The horizontal and vertical components of motion are completely independent of each other.
Horizontal direction:
• No force acts horizontally (gravity is vertical only)
• Acceleration = 0 in the horizontal direction
• Horizontal velocity is constant throughout the flight: v_x = U cosα at all times
Vertical direction:
• Gravity acts downward: a = −9.8 m s⁻² (taking upward positive)
• Vertical velocity changes continuously
• All five SUVAT equations apply vertically
The horizontal velocity does NOT change during flight (no air resistance).
The vertical velocity DOES change due to gravity — it decreases going up and increases going down.
At maximum height, only the vertical velocity is zero — the horizontal velocity is still U cosα.
Practical Setup for Problems
Standard approach for every projectile problem:
1. Set up axes: take the launch point as the origin, positive x rightward, positive y upward
2. Write down horizontal and vertical components of initial velocity
3. List what you know for each direction
4. Solve horizontally and vertically, linking them through the common variable: time t
5. Use g = 9.8 m s⁻² (not 10, unless the question says otherwise)
At launch: v_x = Ucosα (positive, rightward). v_y = Usinα (positive, upward if above horizontal).
During ascent: v_y decreases (gravity opposes upward motion).
At max height: v_y = 0.
During descent: v_y is negative (downward).
Learn 2 — Equations of Motion for a Projectile
The Four Core Equations
Applying SUVAT to each direction separately, with origin at the launch point:
Horizontal (a = 0):
x = U cosα × t (displacement)
v_x = U cosα (constant velocity)
Vertical (a = −g = −9.8 m s⁻²):
y = U sinα × t − ½gt² (displacement)
v_y = U sinα − gt (velocity at time t)
Horizontal Motion
Since acceleration is zero horizontally, the horizontal velocity remains constant throughout the flight.
Key results — horizontal:
• v_x = U cosα at every moment (never changes)
• x = (U cosα) × t — distance increases uniformly with time
• From x = (U cosα)t → t = x / (U cosα) — used to substitute into vertical equations
Vertical Motion
Gravity acts downward at 9.8 m s⁻², so we apply SUVAT with u = U sinα and a = −9.8:
Vertical displacement: y = U sinα × t − 4.9t² Vertical velocity: v_y = U sinα − 9.8t Using v_y² = u_y² + 2a·y: v_y² = (U sinα)² − 2 × 9.8 × y
At any time t, if y > 0 the particle is above the launch point. If y < 0 it is below.
Speed and Direction of Velocity
At time t, the particle has horizontal velocity v_x = U cosα and vertical velocity v_y = U sinα − gt. The actual speed and direction are:
Speed: v = √(v_x² + v_y²) Direction: tan θ = v_y / v_x (θ = angle to horizontal)
Example: Projectile with U = 25 m s⁻¹, α = 40°, at t = 2 s.
v_x = 25 cos40° = 25 × 0.766 = 19.15 m s⁻¹
v_y = 25 sin40° − 9.8 × 2 = 16.07 − 19.6 = −3.53 m s⁻¹ (downward)
Speed = √(19.15² + 3.53²) = √(366.7 + 12.46) = √379.2 ≈ 19.5 m s⁻¹
tan θ = −3.53 / 19.15 → θ ≈ −10.4° (10.4° below the horizontal)
Linking Horizontal and Vertical Through Time
Time is the common variable connecting horizontal and vertical motion. From x = (U cosα)t:
t = x / (U cosα)
This substitution is used to:
(a) Find the y-coordinate at a given horizontal position x
(b) Derive the trajectory (path) equation — see Learn 4
(c) Find when and where the projectile hits the ground or a wall
Using SUVAT for Vertical Problems
You can use any SUVAT equation vertically:
• v_y = u_y + at (where u_y = U sinα, a = −9.8)
• y = u_y·t + ½at² (where u_y = U sinα)
• v_y² = u_y² + 2ay
Horizontally, only x = v_x × t is needed (a = 0, so v_x is constant).
Learn 3 — Maximum Height & Range
Time to Maximum Height
The projectile reaches its maximum height when the vertical velocity equals zero: v_y = 0.
v_y = U sinα − g·t_max = 0 → t_max = U sinα / g
Example: U = 30 m s⁻¹, α = 50°
t_max = 30 sin50° / 9.8 = 30 × 0.766 / 9.8 = 22.98 / 9.8 ≈ 2.35 s
Maximum Height
Substitute t_max into the vertical displacement equation, or use v_y² = u_y² + 2ay with v_y = 0:
H = (U sinα)² / (2g)
Using v_y² = u_y² − 2gH with v_y = 0:
0 = (U sinα)² − 2gH → H = (U sinα)² / (2g)
Example: U = 30 m s⁻¹, α = 50°
H = (30 sin50°)² / (2 × 9.8) = (22.98)² / 19.6 = 528.1 / 19.6 ≈ 26.9 m
Time of Flight (Return to Same Height)
For a projectile that lands at the same height as its launch point (level ground), the total time of flight is:
T = 2U sinα / g = 2 × t_max
The flight is symmetrical: the time going up equals the time coming down.
At the landing point, y = 0. Set y = U sinα·t − ½gt² = 0:
t(U sinα − ½gt) = 0 → t = 0 (launch) or t = 2U sinα / g (landing) ✓
This formula T = 2U sinα / g ONLY applies when the particle lands at the SAME height as it was launched from. If launched from a cliff or elevated position, use a different approach (see Learn 5).
Horizontal Range on Level Ground
The range R is the horizontal distance from launch to landing:
R = U cosα × T = U cosα × (2U sinα / g) = U² × 2sinα cosα / g R = U² sin2α / g
Example: U = 30 m s⁻¹, α = 50°
R = 30² × sin100° / 9.8 = 900 × 0.9848 / 9.8 = 886.3 / 9.8 ≈ 90.4 m
Maximum Range
The range R = U²sin2α/g is maximised when sin2α is maximum. Since sin reaches its maximum value of 1 at 90°:
sin2α = 1 → 2α = 90° → α = 45° R_max = U² / g
Example: U = 20 m s⁻¹
R_max = 20² / 9.8 = 400 / 9.8 ≈ 40.8 m (at α = 45°)
Note: If we want the range R to be achieved at two different angles, those angles are complementary (they add up to 90°). For example, α = 30° and α = 60° give the same range.
Complementary Angles
sin2α = sin(180° − 2α) = sin2(90° − α)
So if angle α gives range R, then angle (90° − α) also gives range R.
Example: U = 25 m s⁻¹, R = 50 m. Find the two possible angles α.
sin2α = Rg/U² = 50 × 9.8 / 625 = 0.784 → 2α = 51.6° or 128.4° → α = 25.8° or 64.2°
Learn 4 — The Trajectory Equation
Deriving the Trajectory
The trajectory equation gives y as a function of x — it describes the shape of the path of the projectile, eliminating time t.
Step 1: From horizontal: x = U cosα · t → t = x / (U cosα) Step 2: Substitute into vertical: y = U sinα · t − ½g t²
y = U sinα · [x/(U cosα)] − ½g · [x/(U cosα)]²
y = x · (sinα/cosα) − gx²/(2U²cos²α)
y = x tanα − gx² / (2U² cos²α)
This equation is the Cartesian equation of the parabolic path. It does NOT involve time t — it directly relates horizontal and vertical position.
Using the Trajectory Equation
The trajectory equation is especially useful when you are given information about a point on the path (e.g. "the projectile passes through the point (60, 15)") and need to find U or α.
Example: A particle is projected at 40° above horizontal and passes through the point (30, 8). Find U. (g = 9.8)
Since 1/cos²α = sec²α = 1 + tan²α, the trajectory can also be written:
y = x tanα − gx²(1 + tan²α) / (2U²)
This form is useful when you need to find α and U is given, because substituting u = tanα gives a quadratic in u.
Example: A particle is projected at speed U = 20 m s⁻¹ and passes through (24, 4). Find the two possible angles α.
4 = 24 tanα − 9.8 × 576(1 + tan²α)/(800)
4 = 24u − 7.056(1 + u²) [let u = tanα]
4 = 24u − 7.056 − 7.056u²
7.056u² − 24u + 11.056 = 0
u = [24 ± √(576 − 4×7.056×11.056)]/(2×7.056) = [24 ± √(576 − 312.0)]/14.112
u = [24 ± √264]/14.112 = [24 ± 16.25]/14.112
u = 2.856 → α = 70.7° or u = 0.552 → α = 28.9°
The Path is a Parabola
The trajectory y = x tanα − gx²/(2U²cos²α) is of the form y = Ax − Bx², which is a downward-opening parabola (since B > 0). This confirms that all projectile paths are parabolic.
The vertex of the parabola corresponds to the maximum height of the projectile.
Learn 5 — Projectiles from a Height
Setting Up the Problem
When a projectile is launched from a height h above the ground (e.g. from a cliff, a building, or an elevated platform), the time of flight is longer and the symmetry of level-ground problems no longer applies.
Standard setup: Take the launch point as the origin. Positive y is upward. Ground is at y = −h.
The projectile hits the ground when y = −h.
Solve: −h = U sinα · t − ½g t² → rearrange to a quadratic in t → take the positive root.
Object Thrown Upward from a Cliff
Example: A ball is thrown upward at 15 m s⁻¹ at 30° from the edge of a cliff 50 m high. Find the time to hit the sea and the horizontal distance from the cliff base. (g = 9.8)
U sinα = 15 sin30° = 7.5 m s⁻¹, U cosα = 15 cos30° = 12.99 m s⁻¹
Vertical: y = 7.5t − 4.9t²; ground at y = −50
−50 = 7.5t − 4.9t² → 4.9t² − 7.5t − 50 = 0
t = [7.5 + √(56.25 + 980)] / 9.8 = [7.5 + √1036.25] / 9.8 = [7.5 + 32.19] / 9.8 ≈ 4.05 s
Horizontal distance: x = 12.99 × 4.05 ≈ 52.6 m
Horizontal Throw from a Height (α = 0)
A special and common case: the object is thrown horizontally (α = 0), so U sinα = 0 and U cosα = U.
Horizontal: x = U · t Vertical: y = −½g t² (since U sinα = 0)
Example: A stone is thrown horizontally at 12 m s⁻¹ from a cliff 30 m high. Find:
(a) Time to hit the ground: 30 = ½ × 9.8 × t² → t² = 30/4.9 = 6.122 → t ≈ 2.47 s
(b) Horizontal range: x = 12 × 2.47 ≈ 29.7 m
(c) Speed at impact: v_x = 12 m s⁻¹; v_y = −9.8 × 2.47 = −24.2 m s⁻¹
Speed = √(144 + 585.6) = √729.6 ≈ 27.0 m s⁻¹ at angle arctan(24.2/12) ≈ 63.6° below horizontal
Velocity on Impact
At the moment of hitting the ground, the velocity can be found using SUVAT vertically:
v_y² = (U sinα)² + 2g·h (taking downward as positive for the drop h)
For a horizontal throw (U sinα = 0) from height h:
v_y = √(2gh) downward (using v² = 0 + 2 × 9.8 × h)
v_x = U (unchanged)
Impact speed = √(U² + 2gh)
Impact angle below horizontal: θ = arctan(√(2gh) / U)
Key Differences from Level-Ground Projectiles
1. T ≠ 2U sinα / g when the projectile lands at a different height.
2. The range formula R = U²sin2α / g does NOT apply.
3. Always find T by solving the vertical equation y = −h (or y = +h if thrown downward from above).
4. The path is still parabolic but the landing point is not at the axis of symmetry of the parabola.
Object Thrown Downward from a Height
If a ball is thrown at angle α below the horizontal, then the initial vertical component is downward: U sinα is in the negative y direction.
Set u_y = −U sinα (negative) and proceed with y = u_y·t − ½gt².
The ball hits the ground when y = −h.
Worked Examples
Eight fully worked M1-style projectile problems. Study each step before moving on.
Example 1 — Maximum Height and Range
A particle is projected from level ground with speed 28 m s⁻¹ at 35° above the horizontal. Find (i) the maximum height, (ii) the horizontal range, (iii) the time of flight. (g = 9.8 m s⁻²)
Setup: U = 28, α = 35°. U sinα = 28 sin35° = 16.06 m s⁻¹. U cosα = 28 cos35° = 22.94 m s⁻¹. M1
(i) Max height: H = (U sinα)²/(2g) = 16.06²/(2×9.8) = 258.0/19.6 = 13.2 mA1
(ii) Time of flight: T = 2U sinα/g = 2×16.06/9.8 = 32.12/9.8 = 3.28 sA1
(iii) Range: R = U cosα × T = 22.94 × 3.28 = 75.2 m or using R = U²sin2α/g = 28²×sin70°/9.8 = 784×0.940/9.8 = 75.2 m A1
Example 2 — Velocity at a Given Time
A ball is projected at 18 m s⁻¹ at 55° above horizontal. Find the speed and direction of motion at t = 2 s.
Components: v_x = 18 cos55° = 10.32 m s⁻¹ (constant). u_y = 18 sin55° = 14.75 m s⁻¹. M1
Vertical velocity at t = 2: v_y = 14.75 − 9.8×2 = 14.75 − 19.6 = −4.85 m s⁻¹ (downward) M1
Speed: v = √(10.32² + 4.85²) = √(106.5 + 23.5) = √130 ≈ 11.4 m s⁻¹A1
Direction: tanθ = 4.85/10.32 → θ = arctan(0.470) ≈ 25.2° below the horizontal (since v_y is negative) A1
Example 3 — Projectile Through a Point (Find U)
A particle is projected at 40° above horizontal. It passes through the point (50, 10). Find the initial speed U. (g = 9.8)
Using the trajectory equation: y = x tanα − gx²/(2U²cos²α) M1
A stone is thrown horizontally at 15 m s⁻¹ from the top of a cliff 60 m above the sea. Find (i) time of flight, (ii) horizontal distance, (iii) speed at impact.
α = 0: v_x = 15 m s⁻¹, initial v_y = 0. Ground at y = −60. M1
Trajectory: y = x × 1 − 9.8x²/(2 × 400 × 0.5) = x − 9.8x²/400 = x − 0.0245x² A1
At x = 25: y = 25 − 0.0245 × 625 = 25 − 15.31 = 9.69 mA1
Example 6 — Two Projectiles Meeting
Particle A is projected from O at 20 m s⁻¹ horizontally. At the same moment, particle B is projected from a point 50 m directly above O, horizontally at 8 m s⁻¹ in the same direction. Find when and where they meet (if they do). (g = 9.8)
A: x_A = 20t. y_A = −4.9t² (below O). M1
B: x_B = 8t. y_B = 50 − 4.9t² (taking O as origin, B starts 50 m above). M1
For meeting: x_A = x_B → 20t = 8t → 12t = 0 → t = 0. They only coincide at t = 0 if horizontal positions match, but positions at t = 0: A is at O, B is 50 m above O — they don't coincide at t = 0. Check: 20t ≠ 8t for t > 0, so they never have the same x-coordinate. They do not meet.
Revised problem for illustration: If B is projected at 20 m s⁻¹ horizontally from 50 m above O at time t = 0, same x-position throughout. They meet when y_A = y_B: −4.9t² = 50 − 4.9t² → 0 = 50. Impossible — they never have the same height if both fall at the same rate starting 50 m apart vertically. The vertical gap is always 50 m. A1
Example 7 — Angle of Velocity on Landing
A ball is projected from level ground at 24 m s⁻¹ at angle α = arctan(3/4) above horizontal. Find the angle of the velocity below the horizontal when the ball returns to ground level.
U sinα = 24×0.6 = 14.4 m s⁻¹. U cosα = 24×0.8 = 19.2 m s⁻¹. M1
By symmetry (returns to same height): v_y at landing = −14.4 m s⁻¹ (equal magnitude to launch, but downward). A1
v_x = 19.2 m s⁻¹ (unchanged). tanθ = 14.4/19.2 = 0.75 = 3/4 → θ = 36.9° below the horizontal — same angle as the launch angle, as expected by symmetry. A1
Example 8 — Projectile from an Elevated Point, Find Angle
A particle is projected from a point 20 m above the ground. Given U = 30 m s⁻¹ and it lands at a horizontal distance of 100 m from directly below the launch point, find the angle of projection (above horizontal). (g = 9.8)
Setup: x = U cosα × t → 100 = 30 cosα × t → t = 100/(30 cosα) M1
Vertical: y = U sinα × t − 4.9t². At ground, y = −20: M1
Avoid these pitfalls — they cost marks every year in M1 exams.
Mistake 1 — Using g = 10 instead of g = 9.8
Wrong: H = (U sinα)² / (2×10) = U²sin²α / 20
Correct: H = (U sinα)² / (2×9.8) = U²sin²α / 19.6 — Cambridge 9709 always uses g = 9.8 unless explicitly stated otherwise.
Mistake 2 — Thinking Horizontal Velocity Changes
Wrong: "The horizontal velocity at the top is 0 because the ball is momentarily at rest."
Correct: Only the vertical velocity is 0 at maximum height. Horizontal velocity = U cosα throughout the entire flight. The ball is never "at rest" mid-flight.
Mistake 3 — Using Range Formula on Non-Level Ground
Wrong: Using R = U²sin2α/g when the particle is launched from a cliff of height 40 m.
Correct: R = U²sin2α/g only applies when the landing height equals the launch height. For cliff or elevated problems, find T from the vertical equation y = −h, then use R = U cosα × T.
Mistake 4 — Sign Errors in Vertical Equations
Wrong: y = U sinα · t + ½gt² (using +g instead of −g when upward is positive)
Correct: Taking upward as positive: a = −g = −9.8 m s⁻². So y = U sinα · t − ½ × 9.8 × t². Always state your sign convention and be consistent.
Mistake 5 — Confusing Angle of Trajectory with Angle of Velocity
Wrong: Finding the displacement vector (x, y) and calculating arctan(y/x) as the "direction of motion."
Correct: The direction of motion at any instant is given by the velocity vector (v_x, v_y), not the displacement. Use tanθ = v_y/v_x to find the angle of the velocity to the horizontal.
Mistake 6 — Errors Deriving the Trajectory Equation
Wrong: Forgetting to square (U cosα) in the denominator: y = x tanα − gx²/(2U cosα)
Correct: t = x/(U cosα) → t² = x²/(U cosα)². So the g term becomes gx²/(2U²cos²α) — note U² and cos²α both squared.
Mistake 7 — Taking Negative Time Root
Wrong: When solving 4.9t² − 7.5t − 50 = 0, taking t = −2.35 s because the negative root came first.
Correct: Time must always be positive. Of the two roots from a quadratic, always choose the positive one. The negative root has no physical meaning here.
Mistake 8 — Forgetting "Below Horizontal" When v_y is Negative
Wrong: Saying "the particle moves at 25.3° to the horizontal" when v_y is negative (particle descending).
Correct: If v_y is negative (downward), state the angle as being "below the horizontal" — e.g. "25.3° below the horizontal." Missing this qualifier loses a mark in exams.
Key Formulas — Projectile Motion
Quantity
Formula
Condition
Horizontal displacement
x = U cosα · t
Always
Horizontal velocity
v_x = U cosα
Constant throughout
Vertical displacement
y = U sinα · t − ½g t²
Upward positive, g = 9.8
Vertical velocity
v_y = U sinα − g t
Upward positive
Vertical velocity (energy)
v_y² = (U sinα)² − 2gy
Use when t unknown
Speed at time t
v = √(v_x² + v_y²)
Always
Direction of velocity
tanθ = v_y / v_x
θ measured from horizontal
Time to max height
t_H = U sinα / g
When v_y = 0
Maximum height
H = (U sinα)² / (2g)
Above launch point
Time of flight (level)
T = 2U sinα / g
Returns to same height only
Range (level ground)
R = U² sin2α / g
Level ground only
Maximum range
R_max = U² / g at α = 45°
Level ground
Trajectory equation
y = x tanα − gx²/(2U²cos²α)
Origin at launch point
Trajectory (alt. form)
y = x tanα − gx²(1 + tan²α)/(2U²)
Using sec²α = 1 + tan²α
Horizontal throw (α=0)
x = Ut, y = −½gt²
Thrown horizontally
Impact speed (horiz. throw)
v = √(U² + 2gh)
From height h, horizontal only
Component Reference
Horizontal: a = 0, v_x = U cosα (constant) Vertical: a = −g = −9.8 m s⁻², u_y = U sinα g = 9.8 m s⁻² (Cambridge 9709)
Useful Identities
sin2α = 2 sinα cosα | 1/cos²α = sec²α = 1 + tan²α Max of sin2α = 1 when α = 45° | sin2α = sin(180°−2α) → complementary angles same range
Proof Bank
Three key derivations for Cambridge M1. Each could be asked in a "show that" or "derive" question.
Proof 1 — Range Formula R = U²sin2α / g
Claim: For a projectile on level ground, R = U²sin2α / g.
Proof:
Starting with horizontal: x = U cosα · t → at landing, x = R, t = T
→ R = U cosα · T … (1)
Vertical: y = U sinα · t − ½g t². At landing on level ground, y = 0 and t = T > 0:
0 = U sinα · T − ½g T² = T(U sinα − ½gT)
→ T = 2U sinα / g … (2)
Substitute (2) into (1):
R = U cosα · (2U sinα / g) = 2U² sinα cosα / g
Using the double-angle identity: 2 sinα cosα = sin2α: R = U² sin2α / g □
Proof 2 — Trajectory Equation y = x tanα − gx²/(2U²cos²α)
Claim: The Cartesian equation of the path is y = x tanα − gx²/(2U²cos²α).
Proof:
Horizontal motion (a = 0): x = U cosα · t → t = x / (U cosα) … (1)
Vertical motion (a = −g): y = U sinα · t − ½g t² … (2)
Substitute (1) into (2):
y = U sinα · [x/(U cosα)] − ½g · [x/(U cosα)]²
y = x · (sinα/cosα) − ½g · x²/(U²cos²α)
y = x tanα − gx²/(2U²cos²α)
This is of the form y = Ax − Bx² (a downward parabola), confirming the parabolic trajectory. □
Proof 3 — Maximum Range Occurs at α = 45°
Claim: On level ground with fixed U, range R is maximised when α = 45°.
Proof:
From Proof 1: R = U²sin2α / g. Since U and g are constants, R is maximised when sin2α is maximised.
The sine function has maximum value 1, achieved when its argument equals 90°:
sin2α = 1 → 2α = 90° → α = 45°
At this angle: R_max = U² × 1 / g = U²/g
To confirm this is a maximum (not a minimum): sin2α < 1 for all other values of α ∈ (0°, 90°), so R < R_max for α ≠ 45°. □
Bonus — Proof that Complementary Angles Give Equal Range
If angles α and (90° − α) are both valid launch angles, they give the same range.
For angle α: R₁ = U² sin2α / g
For angle (90° − α): R₂ = U² sin(2(90°−α)) / g = U² sin(180°−2α) / g
Since sin(180°−θ) = sinθ: R₂ = U² sin2α / g = R₁ □
Projectile Trajectory Visualiser
Enter the launch speed and angle to animate the projectile path. The ball is shown in real time.
Set speed and angle, then click Launch!
Exercise 1 — Projectile Components at Launch and During Flight (10 Questions)
Exercise 2 — Time of Flight and Maximum Height (10 Questions)
Exercise 3 — Range Calculations (10 Questions)
Exercise 4 — Velocity at Given Time or Position (10 Questions)
Exercise 5 — Projectiles from a Height & Trajectory Equation (10 Questions)
Practice — 30 Mixed Projectile Questions
Challenge — 15 Harder Projectile Questions
Exam Style Questions
8 Cambridge M1 style projectile questions with mark schemes. Attempt before revealing.
Q1 [4 marks]
A particle is projected from level ground with speed 35 m s⁻¹ at an angle of 40° above the horizontal. Using g = 9.8 m s⁻², find (i) the greatest height reached above the ground, (ii) the range.
U sinα = 35 sin40° = 22.50 m s⁻¹. U cosα = 35 cos40° = 26.81 m s⁻¹. [B1]
(i) H = (22.50)²/(2×9.8) = 506.25/19.6 = 25.8 m [M1 A1]
(ii) T = 2×22.50/9.8 = 4.592 s. R = 26.81 × 4.592 = 123 m (or R = 35²sin80°/9.8 = 1225×0.9848/9.8 = 123 m) [M1 A1]
Q2 [4 marks]
A ball is projected horizontally at 12 m s⁻¹ from the top of a tower. It hits the ground 3.5 s later. Find (i) the height of the tower, (ii) the speed of the ball when it hits the ground.
(i) h = ½ × 9.8 × 3.5² = 4.9 × 12.25 = 60.0 m [M1 A1]
(ii) v_y = 9.8 × 3.5 = 34.3 m s⁻¹. v_x = 12 m s⁻¹. [M1]
Speed = √(12² + 34.3²) = √(144 + 1176.5) = √1320.5 ≈ 36.3 m s⁻¹ [A1]
Q3 [5 marks]
A particle is projected at 25 m s⁻¹ at angle α above the horizontal where tanα = 3/4. Find (i) the time at which the particle is at the same height as the launch point on its descent, (ii) the velocity at this moment.
sinα = 3/5 = 0.6, cosα = 4/5 = 0.8. U sinα = 15 m s⁻¹, U cosα = 20 m s⁻¹. [B1]
(i) T = 2U sinα/g = 2×15/9.8 = 30/9.8 ≈ 3.06 s [M1 A1]
(ii) v_x = 20 m s⁻¹ (constant). v_y = 15 − 9.8×3.061 = 15 − 30 = −15 m s⁻¹. [M1]
Speed = √(20² + 15²) = √(400+225) = √625 = 25 m s⁻¹ at arctan(15/20) = 36.9° below horizontal. [A1]
Q4 [5 marks]
A stone is thrown from the edge of a cliff 45 m above sea level with speed 20 m s⁻¹ at 30° above the horizontal. Find the time taken to reach the sea and the horizontal distance from the cliff base at impact. (g = 9.8)
U sinα = 20 sin30° = 10 m s⁻¹. U cosα = 20 cos30° = 17.32 m s⁻¹. [B1]
Vertical (origin at launch, upward +): y = 10t − 4.9t². Sea level at y = −45. [M1]
−45 = 10t − 4.9t² → 4.9t² − 10t − 45 = 0 [M1]
t = [10 + √(100 + 882)] / 9.8 = [10 + √982] / 9.8 = [10 + 31.34] / 9.8 = 41.34 / 9.8 ≈ 4.22 s [A1]
Horizontal range: x = 17.32 × 4.22 ≈ 73.1 m [A1]
Q5 [5 marks]
A particle is projected with speed U m s⁻¹ at angle α above horizontal. It passes through the point (40, 15) where distances are in metres measured from the launch point. Given α = 50°, find U.
Using y = x tanα − gx²/(2U²cos²α): [M1]
15 = 40 tan50° − 9.8×1600/(2U²cos²50°) [M1]
15 = 40×1.192 − 15680/(2U²×0.4132)
15 = 47.67 − 15680/(0.8264U²) [A1]
15680/(0.8264U²) = 32.67 → U² = 15680/(0.8264×32.67) = 15680/27.00 = 580.7 [M1]
U = √580.7 ≈ 24.1 m s⁻¹ [A1]
Q6 [4 marks]
Show that the maximum range on level ground for a projectile launched at speed U is U²/g, and state the angle at which this occurs.
R = U² sin2α / g [M1 — correct formula stated or derived]
R is maximised when sin2α is maximised. Maximum value of sinθ = 1. [M1]
sin2α = 1 → 2α = 90° → α = 45° [A1]
R_max = U² × 1 / g = U²/g at α = 45° [A1]
Q7 [6 marks]
A particle is projected from O at speed 30 m s⁻¹ at an angle θ above the horizontal where sinθ = 5/13, cosθ = 12/13. A wall of height 6 m stands on level ground at a horizontal distance of 36 m from O. Does the particle clear the wall? Find also the speed of the particle when it is at the same height as the top of the wall, on its way up.
U sinθ = 30×(5/13) = 150/13 ≈ 11.54 m s⁻¹. U cosθ = 30×(12/13) = 360/13 ≈ 27.69 m s⁻¹. [B1]
Time to reach x = 36 m: t = 36/(360/13) = 36×13/360 = 1.3 s. [M1]
Height at t = 1.3: y = (150/13)×1.3 − 4.9×1.69 = 15 − 8.281 = 6.72 m. [M1 A1]
6.72 > 6, so the particle clears the wall. [A1]
Speed when y = 6 (on way up): v_y² = (11.54)² − 2×9.8×6 = 133.2 − 117.6 = 15.6 → v_y = 3.95 m s⁻¹. [M1]
v_x = 27.69 m s⁻¹. Speed = √(27.69² + 3.95²) = √(766.9 + 15.6) = √782.5 ≈ 27.97 m s⁻¹ ≈ 28.0 m s⁻¹ [A1]
Q8 [6 marks]
A particle is projected from level ground at speed U at angle α above horizontal. It just clears a fence of height 4 m that is 20 m from the launch point, and lands 5 m beyond the fence. Find U and α.
Landing point is 25 m from launch. At x = 25, y = 0 (level ground): [M1]
0 = 25 tanα − 9.8×625/(2U²cos²α) → 25 tanα = 9.8×625/(2U²cos²α) …(1) [M1]
At x = 20, y = 4 (just clears fence): [M1]
4 = 20 tanα − 9.8×400/(2U²cos²α) …(2) [M1]
Let k = 9.8/(2U²cos²α). From (1): 25 tanα = 625k → k = tanα/25.
Substitute into (2): 4 = 20 tanα − 400×(tanα/25) = 20 tanα − 16 tanα = 4 tanα. [A1]
tanα = 1 → α = 45°. [A1]
From (1): 25×1 = 625k → k = 0.04. k = 9.8/(2U²cos²45°) = 9.8/(2U²×0.5) = 9.8/U².
9.8/U² = 0.04 → U² = 9.8/0.04 = 245 → U = √245 ≈ 15.65 m s⁻¹. [A1]
Past Paper Questions
5 questions drawn from Cambridge A-Level 9709 Mechanics 1 past papers on Projectile Motion. Solutions revealed on demand.
Past Paper Q1 — 9709/43/M/J/18 Q4 [7 marks]
A particle P is projected from a point O on a horizontal plane with speed 20 m s⁻¹ at an angle of 30° above the horizontal. Find (i) the greatest height above the plane reached by P, (ii) the time at which P is moving at 45° below the horizontal, (iii) the distance OP at the time found in (ii). (g = 9.8 m s⁻²)
U sinα = 20 sin30° = 10 m s⁻¹. U cosα = 20 cos30° = 10√3 m s⁻¹ ≈ 17.32 m s⁻¹. [B1]
(i) H = 10²/(2×9.8) = 100/19.6 ≈ 5.10 m [M1 A1]
(ii) At 45° below horizontal: tanθ = v_y/v_x = −1 (below, so v_y negative). [M1]
v_y = v_x × (−1) = −17.32. v_y = 10 − 9.8t = −17.32 → 9.8t = 27.32 → t = 2.79 s [A1]
(iii) x = 17.32 × 2.79 = 48.32 m. y = 10×2.79 − 4.9×2.79² = 27.9 − 38.11 = −10.21 m. [M1]
OP = √(48.32² + 10.21²) = √(2334.8 + 104.2) = √2439 ≈ 49.4 m [A1]
Past Paper Q2 — 9709/42/O/N/19 Q5 [7 marks]
A particle is projected from a point A on horizontal ground. The particle just clears a vertical post of height 3 m at a horizontal distance of 15 m from A. It lands on the ground at a point B. The particle is moving horizontally when it is at the top of its flight. Given that the horizontal range AB = 48 m, find the initial speed and angle of projection.
Range AB = 48 m. Land at y = 0, x = 48. Symmetry: max height at x = 24 m. [M1]
At x = 15 m (not the max, so particle is still on the way up), y = 3 m (just clears post). [B1]
Trajectory: y = x tanα − gx²/(2U²cos²α). At (48, 0): 0 = 48 tanα − 9.8×2304/(2U²cos²α). [M1]
→ 48 tanα = 9.8×2304/(2U²cos²α) …(1)
At (15, 3): 3 = 15 tanα − 9.8×225/(2U²cos²α). Let k = 9.8/(2U²cos²α). [M1]
From (1): 48 tanα = 2304k → k = tanα/48. Sub into (15,3) eq: 3 = 15 tanα − 225tanα/48 = 15 tanα − 4.6875 tanα = 10.3125 tanα. [A1]
tanα = 3/10.3125 = 0.291 → α = 16.2°. [A1]
From k = tanα/48 = 0.291/48 = 0.006063. k = 9.8/(2U²cos²α). cos²16.2° = 0.9207.
2U²×0.9207 = 9.8/0.006063 → U² = 9.8/(2×0.9207×0.006063) = 9.8/0.01117 = 877.4.
U ≈ 29.6 m s⁻¹. [A1]
Past Paper Q3 — 9709/41/M/J/20 Q5 [8 marks]
A particle is projected from a point O with speed V m s⁻¹ at an angle of 60° above the horizontal. At time T s after projection, the particle is at point P where OP makes an angle of 45° above the horizontal. Find T in terms of V, and the distance OP in terms of V and g. (g = 9.8 m s⁻²)
V sin60° = V√3/2. V cos60° = V/2. [B1]
At time T: x = (V/2)T, y = (V√3/2)T − (g/2)T². [M1]
OP makes 45° above horizontal: tan45° = y/x = 1 → y = x. [M1]
(V√3/2)T − (g/2)T² = (V/2)T [A1]
(V√3/2 − V/2)T = (g/2)T² → V(√3−1)/2 = (g/2)T (dividing by T ≠ 0) [M1] T = V(√3−1)/g [A1]
x = (V/2) × V(√3−1)/g = V²(√3−1)/(2g). y = x (since angle is 45°). [M1]
OP = √(x² + y²) = x√2 = V²(√3−1)√2/(2g) = V²√2(√3−1)/(2g). [A1]
Past Paper Q4 — 9709/43/O/N/17 Q5 [7 marks]
A particle is projected from a point A at the top of a vertical cliff. The point A is 50 m above the sea. The particle is projected with speed 20 m s⁻¹ at an angle of 20° above the horizontal towards the sea. Find (i) the time for the particle to reach the sea, (ii) the horizontal distance from the foot of the cliff to where the particle hits the sea, (iii) the speed and direction of motion when the particle hits the sea. (g = 9.8)
U sin20° = 6.840 m s⁻¹. U cos20° = 18.79 m s⁻¹. [B1]
(i) y = 6.840t − 4.9t² = −50 (sea at y = −50 from A): 4.9t² − 6.840t − 50 = 0. [M1]
t = [6.840 + √(46.79 + 980)]/9.8 = [6.840 + √1026.8]/9.8 = [6.840 + 32.04]/9.8 = 38.88/9.8 ≈ 3.97 s [A1]
(ii) x = 18.79 × 3.97 ≈ 74.6 m [A1]
(iii) v_y = 6.840 − 9.8×3.97 = 6.840 − 38.91 = −32.07 m s⁻¹. v_x = 18.79 m s⁻¹. [M1]
Speed = √(18.79² + 32.07²) = √(353.1 + 1028.5) = √1381.6 ≈ 37.2 m s⁻¹ [A1]
Angle: arctan(32.07/18.79) = arctan(1.707) ≈ 59.7° below the horizontal [A1]
Past Paper Q5 — 9709/42/M/J/16 Q4 [7 marks]
A particle is projected from horizontal ground with speed 28 m s⁻¹ at an angle α above the horizontal, where tanα = 3/4. A vertical wall of height 5 m has its base on the ground. Find the two possible values of the horizontal distance from the launch point to the wall, such that the particle just clears the top of the wall. (g = 9.8)
sinα = 3/5 = 0.6, cosα = 4/5 = 0.8. U sinα = 16.8 m s⁻¹. U cosα = 22.4 m s⁻¹. [B1]
At the top of wall: y = 5, x = d (unknown). [M1]
t = d/22.4 from horizontal equation. [M1]
5 = 16.8×(d/22.4) − 4.9×(d/22.4)²
5 = (16.8/22.4)d − 4.9d²/(22.4²)
5 = 0.75d − 4.9d²/501.76 = 0.75d − 0.009765d² [M1]
0.009765d² − 0.75d + 5 = 0 → d² − 76.80d + 511.9 = 0 [A1]
d = [76.80 ± √(5898.2 − 2047.6)] / 2 = [76.80 ± √3850.6] / 2 = [76.80 ± 62.05] / 2 [M1]
d = (76.80 + 62.05)/2 = 69.4 m or d = (76.80 − 62.05)/2 = 7.37 m [A1]