Projectile Motion

Modelling objects moving under gravity with horizontal and vertical components

SPEC 7.5 • KINEMATICS

What is Projectile Motion?

Definition

Projectile motion is the motion of an object that moves through the air under the influence of gravity alone, with an initial velocity that has both horizontal and vertical components.

Key Assumptions

Air resistance is negligible (no drag)

The only force acting is gravity (downwards)

Acceleration due to gravity g = 9.8 m/s² (constant)

The projectile is a particle (no rotation)

Key Insight

Horizontal and vertical motions are independent of each other. We can analyse them separately using different equations.

Resolving Initial Velocity

When a projectile is launched with speed u at angle θ to the horizontal, we resolve into components:

Horizontal

u_{x} = u cos θ

Constant throughout flight

Vertical

u_{y} = u sin θ

Changes due to gravity

Remember

If given tan θ = a/b, then: cos θ = b/√(a² + b²) and sin θ = a/√(a² + b²)

Horizontal and Vertical Motion

Horizontal

No acceleration horizontally!

Therefore horizontal velocity is always constant:

v_{x} = u cos θ

Displacement:

x = (u cos θ) t

Vertical

Acceleration = −g (taking up as +ve)

Use SUVAT equations:

v_{y} = u sin θ - g t

y = (u sin θ) t - \frac{1}{2} g t^{2}

Critical Connection

The ONLY thing connecting horizontal and vertical motion is TIME

Both motions happen simultaneously - they share the same value of t

The SUVAT Method

For every projectile problem, list SUVAT separately for horizontal and vertical:

Horizontal

s= x (horizontal distance)u= u cos θv= u cos θ (same - constant!)a= 0t= ?Only use: s = ut (since a = 0)

Vertical

s= y (vertical displacement)u= u sin θv= ? (changes with time)a= −g (or +g if down is +ve)t= ?Use any SUVAT equation

Method

List SUVAT for horizontal AND vertical separately

Identify what you know and what you need

Choose the equation with 3 knowns + 1 unknown

Remember: t is the same in both directions!

Deriving Time of Flight

For a projectile returning to the same level (y = 0):

Derivation

Using:

y = (u sin θ) t - \frac{1}{2} g t^{2}

Set y = 0:

0 = t (u sin θ - \frac{1}{2} g t)

So t = 0 or

t = \frac{2 u s i n θ}{g}

Time of Flight Formula

T = \frac{2 u s i n θ}{g}

Time to return to same level (landing height = launch height)

Deriving Maximum Height

At maximum height, vertical velocity = 0:

Derivation

Using v² = u² + 2as vertically:

0 = (u sin θ)^{2} - 2 g H

2 g H = u^{2} sin^{2} θ

Maximum Height Formula

H = \frac{u ^{2} s i n ^{2} θ}{2 g}

At max height: only horizontal velocity remains (vᵧ = 0)

Deriving Range

Range = horizontal distance when returning to same level:

Derivation

Horizontal:

R = (u cos θ) \times T

Substitute T = (2u sin θ)/g:

R = \frac{2 u ^{2} s i n θ c o s θ}{g}

Using 2 sin θ cos θ = sin 2θ:

Range Formula

R = \frac{u ^{2} s i n 2 θ}{g}

Maximum range when θ = 45° (sin 90° = 1)

Complementary Angles

30° and 60° give the same range (sin 60° = sin 120°). Generally, θ and (90° − θ) give equal range.

Velocity at Any Point

At any time t, velocity has two components:

v_{x} = u cos θ

(constant!)

v_{y} = u sin θ - g t

v = v_{x}^{2} + v_{y}^{2}

tan α = \frac{v _{y}}{v _{x}}

Direction

If vᵧ > 0: moving upward. If vᵧ < 0: moving downward. State angle as "above" or "below" horizontal.

Worked Example01

A ball is kicked with speed 20 m/s at an angle of 30° to the horizontal. Find the horizontal and vertical components of the initial velocity.

1

Given: u = 20 m/s, θ = 30°

2

Horizontal: uₓ = u cos θ = 20 × cos 30° = 20 × (√3/2) = 10√3 m/s ≈ 17.3 m/s

3

Vertical: uᵧ = u sin θ = 20 × sin 30° = 20 × 0.5 = 10 m/s

Worked Example02

A projectile is launched from ground level with speed 25 m/s at 40° to the horizontal. Find: (a) the time of flight, (b) the range.

1

SUVAT (Vertical): s = 0, u = 25 sin 40° = 16.07, v = ?, a = −9.8, t = ?

2

(a) Using T = (2u sin θ)/g = (2 × 25 × sin 40°)/9.8

3

T = (50 × 0.6428)/9.8 = 32.14/9.8 = 3.28 s

4

(b) Range: R = (u² sin 2θ)/g = (625 × sin 80°)/9.8

5

R = (625 × 0.9848)/9.8 = 615.5/9.8 = 62.8 m

Worked Example03

A stone is thrown at 18 m/s at an angle of 60° above the horizontal. Find the maximum height reached.

1

SUVAT (Vertical): s = H, u = 18 sin 60° = 15.59, v = 0 (at max), a = −9.8, t = ?

2

Using v² = u² + 2as: 0 = 15.59² + 2(−9.8)H

3

19.6H = 243 → H = 243/19.6 = 12.4 m

4

Or use formula: H = (u² sin² θ)/(2g) = (324 × 0.75)/19.6 = 12.4 m ✓

Worked Example04

A ball is projected at 30 m/s at 50° to the horizontal. Find the speed and direction of motion after 2 seconds.

1

SUVAT (Vertical): s = ?, u = 30 sin 50° = 22.98, v = ?, a = −9.8, t = 2

2

Horizontal (constant): vₓ = 30 cos 50° = 19.28 m/s

3

Vertical: vᵧ = u + at = 22.98 − 9.8(2) = 22.98 − 19.6 = 3.38 m/s

4

Speed: v = √(19.28² + 3.38²) = √(371.7 + 11.4) = √383.1 = 19.6 m/s

5

Direction: tan α = 3.38/19.28 = 0.175 → α = 9.9° above horizontal

Deriving Equation of Trajectory

We eliminate time t to find the path (trajectory) equation:

Derivation

Step 1: From horizontal:

x = (u cos θ) t

→

t = \frac{x}{u c o s θ}

Step 2: Substitute into

y = (u sin θ) t - \frac{1}{2} g t^{2}

:

y = x tan θ - \frac{g x ^{2}}{2 u ^{2} c o s ^{2} θ}

Step 3: Using identity

sec^{2} θ = \frac{1}{c o s ^{2} θ} = 1 + tan^{2} θ

:

Form 1

y = x tan θ - \frac{g x ^{2}}{2 u ^{2} c o s ^{2} θ}

Form 2

y = x tan θ - \frac{g x ^{2} ( 1 + t a n ^{2} θ )}{2 u ^{2}}

When to Use Each Form

Form 1: When you know the angle θ directly

Form 2: When given tan θ (e.g., tan θ = 3/4) - avoids finding θ first!

Projectile Motion with Vectors

We can describe projectile motion using position and velocity vectors:

Position Vector

r = x i + y j

r = (u cos θ) t i + ((u sin θ) t - \frac{1}{2} g t^{2}) j

Velocity Vector

v = v_{x} i + v_{y} j

v = (u cos θ) i + (u sin θ - g t) j

Using Vector Notation

The acceleration is constant:

a = - g j

(or

a = (0 - g)

)

Vector SUVAT

v = u + a t

r = u t + \frac{1}{2} a t^{2}

Example: u = 20i + 15j m/s from origin

Position:

r = 20 t i + (15 t - 4.9 t^{2}) j

Velocity:

v = 20 i + (15 - 9.8 t) j

Speed:

∣ v ∣ = v_{x}^{2} + v_{y}^{2}

Distance:

∣ r ∣ = x^{2} + y^{2}

Projectiles from a Height

When launched from height h above landing point, set up SUVAT carefully:

Vertical SUVAT

s= −h (below start)u= u sin θv= ?a= −gt= ?

Method

Using s = ut + ½at²:

- h = (u sin θ) t - \frac{1}{2} g t^{2}

Rearrange:

\frac{1}{2} g t^{2} - (u sin θ) t - h = 0

→ Solve using quadratic formula

Thrown Downward

If thrown at angle θ below horizontal: use uᵧ = −u sin θ (negative)

Worked Example05

A stone is thrown horizontally at 12 m/s from the top of a cliff 45 m high. Find: (a) the time to reach the ground, (b) the horizontal distance from the cliff.

1

SUVAT (Vertical, down +ve): s = 45, u = 0, v = ?, a = 9.8, t = ?

2

Horizontal: uₓ = 12 m/s (constant)

3

(a) Using s = ut + ½at²: 45 = 0 + ½(9.8)t² → t² = 9.18 → t = 3.03 s

4

(b) Horizontal distance: x = uₓ × t = 12 × 3.03 = 36.4 m

Worked Example06

A ball is projected from ground level with speed 28 m/s. Find the two possible angles of projection for a range of 60 m.

1

Given: u = 28 m/s, R = 60 m, g = 9.8 m/s², θ = ?

2

Using R = (u² sin 2θ)/g: 60 = (784 × sin 2θ)/9.8

3

sin 2θ = (60 × 9.8)/784 = 588/784 = 0.75

4

2θ = sin⁻¹(0.75) = 48.6° or 180° − 48.6° = 131.4°

5

θ = 24.3° or θ = 65.7°

Worked Example07

A ball is thrown at 20 m/s at 53° above horizontal from ground level. Find the time(s) when the ball is at height 15 m.

1

SUVAT (Vertical): s = 15, u = 20 sin 53° = 16, v = ?, a = −9.8, t = ?

2

Using s = ut + ½at²: 15 = 16t − 4.9t²

3

Rearrange: 4.9t² − 16t + 15 = 0

4

Discriminant = 256 − 4(4.9)(15) = 256 − 294 = −38 < 0

5

No real solutions - the ball never reaches 15 m!

6

Check: Max height = (16)²/(2×9.8) = 13.1 m < 15 m ✓

Practice

Question 1: A football is kicked with speed 24 m/s at an angle of 35° to the horizontal. Find the time of flight and maximum height.

Time of flight: T = (2 × 24 × sin 35°)/9.8 = (48 × 0.574)/9.8 = 27.55/9.8 = 2.81 s

Maximum height: H = (24² × sin² 35°)/(2 × 9.8) = (576 × 0.329)/19.6 = 189.5/19.6 = 9.67 m

Practice

Question 2: A stone is projected from a cliff top 80 m above sea level with speed 15 m/s at 25° above horizontal. Find the time to reach the sea and the horizontal distance travelled.

uₓ = 15 cos 25° = 13.59 m/s, uᵧ = 15 sin 25° = 6.34 m/s

Using y = uᵧt − ½gt² with y = −80 (below start):

−80 = 6.34t − 4.9t² → 4.9t² − 6.34t − 80 = 0

t = (6.34 + √(40.2 + 1568))/9.8 = (6.34 + 40.1)/9.8 = 4.74 s

Horizontal distance: x = 13.59 × 4.74 = 64.4 m

Practice

Question 3: A projectile is launched at 40 m/s at angle θ. The range is 120 m. Find the two possible values of θ.

Using R = (u² sin 2θ)/g:

120 = (1600 × sin 2θ)/9.8

sin 2θ = (120 × 9.8)/1600 = 1176/1600 = 0.735

2θ = sin⁻¹(0.735) = 47.3° or 132.7°

θ = 23.7° or θ = 66.3°

Summary

Key Equations

QuantityFormula

Time of flightT = (2u sin θ)/g

Max heightH = (u² sin² θ)/(2g)

RangeR = (u² sin 2θ)/g

Trajectoryy = x tan θ − gx²/(2u² cos² θ)

Vector Form

Position:

r = (u cos θ)t i + ((u sin θ)t − ½gt²) j

Velocity:

v = (u cos θ) i + (u sin θ − gt) j

Key Exam Skills

List SUVAT separately for horizontal and vertical

Remember: time t is the ONLY connection between H and V

Use trajectory equation when given tan θ (use sec² identity)

Two angles give same range (complementary-type)