Projectile Motion

Analysing the motion of objects moving through uniform gravitational fields

SPEC 3.4.1.4 • MECHANICS AND MATERIALS

The Independence Principle

Key Principle

In a uniform gravitational field, the horizontal and vertical components of a projectile's motion are completely independent of each other.

Horizontal Motion

No horizontal forces (ignoring air resistance)

Therefore: constant velocity

a_{x} = 0

Vertical Motion

Gravitational force acts downward

Therefore: constant acceleration

a_{y} = g = 9.81 m s^{- 2}

Classic Demonstration

Two balls released simultaneously — one dropped, one projected horizontally — hit the ground at the same time. The horizontal velocity has no effect on vertical motion.

Visualising Independence

Key Insight

At any time t, both balls are at the same vertical position. The horizontal motion of the projected ball does not affect its rate of fall.

Equations of Motion

We apply the SUVAT equations separately to each direction:

Horizontal (x)

Since

a_{x} = 0

:

x = u_{x} t

Velocity remains constant:

v_{x} = u_{x}

Vertical (y)

With

a_{y} = g

(taking down as positive):

y = u_{y} t + \frac{1}{2} g t^{2}

v_{y} = u_{y} + g t

v_{y}^{2} = u_{y}^{2} + 2 g y

Sign Convention

Choose a consistent sign convention. Common choices: taking upward as positive (so

a = - g

) or downward as positive (so

a = + g

). State your choice clearly!

Horizontal Launch

When a projectile is launched horizontally from height h:

Initial Conditions

For horizontal launch:

u_{x} = u

and

u_{y} = 0

Direction of motion:

tan θ = \frac{v _{y}}{v _{x}}

where θ is angle below horizontal

Worked Example01

A ball is projected horizontally at 15 m s⁻¹ from the top of a cliff 20 m high. Taking g = 9.81 m s⁻², calculate: (a) the time of flight, (b) the horizontal range.

1

Identify known values:

u_{x} = 15 m s^{- 1}

,

u_{y} = 0

,

h = 20 m

,

g = 9.81 m s^{- 2}

2

(a) Time of flight: Use vertical motion with

s = u_{y} t + \frac{1}{2} g t^{2}

20 = 0 + \frac{1}{2} \times 9.81 \times t^{2}

t^{2} = \frac{20 \times 2}{9.81} = 4.077

t = 2.02 s

3

(b) Horizontal range: Use

x = u_{x} t

R = 15 \times 2.02 = 30.3 m

Answers: (a) t = 2.02 s, (b) R = 30.3 m

Worked Example02

A stone is thrown horizontally at 12 m s⁻¹ from a bridge. It hits the water 1.5 s later. Calculate: (a) the height of the bridge, (b) the speed of the stone on impact.

1

(a) Height: Use

s = u_{y} t + \frac{1}{2} g t^{2}

with

u_{y} = 0

h = 0 + \frac{1}{2} \times 9.81 \times 1. 5^{2} = 11.0 m

2

(b) Find velocity components on impact:

v_{x} = u_{x} = 12 m s^{- 1}

(unchanged)

v_{y} = u_{y} + g t = 0 + 9.81 \times 1.5 = 14.7 m s^{- 1}

3

Resultant speed: Use Pythagoras

v = v_{x}^{2} + v_{y}^{2} = 1 2^{2} + 14. 7^{2} = 144 + 216.1 = 19.0 m s^{- 1}

Answers: (a) h = 11.0 m, (b) v = 19.0 m s⁻¹

Angled Launch

When a projectile is launched at angle θ to the horizontal with initial speed u:

Component Equations

u_{x} = u cos θ u_{y} = u sin θ

Problem-Solving Method

Follow these 4 steps to solve any projectile problem:

Step 1: Resolve Components

Find horizontal info:

s_{x}, u_{x}, v_{x}

and

a_{x} = 0

Find vertical info:

s_{y}, u_{y}, v_{y}

and

a_{y} = g = 9.81 m s^{- 2}

Note: Time t is the same for both components!

Step 2: Choose Equations

Select which SUVAT equation to use for each component. Often different equations for horizontal and vertical.

Step 3: Solve Separately

Solve horizontal and vertical motions independently. Use vertical motion to find time if needed.

Step 4: Recombine

If finding resultant velocity or displacement:

v = v_{x}^{2} + v_{y}^{2} tan θ = \frac{v _{y}}{v _{x}}

Key Results

For a projectile launched from and landing on level ground (taking upward as positive):

QuantityFormulaDerivation

Time to max height

t_{ma x} = \frac{u s i n θ}{g}

Set

v_{y} = 0

in

v_{y} = u_{y} - g t

Total time of flight

T = \frac{2 u s i n θ}{g}

Symmetry:

T = 2 t_{ma x}

Maximum height

H = \frac{u ^{2} s i n ^{2} θ}{2 g}

Use

v_{y}^{2} = u_{y}^{2} - 2 g H

with

v_{y} = 0

Range

R = \frac{u ^{2} s i n 2 θ}{g}

R = u_{x} \times T

Maximum Range

The range is maximum when

sin 2 θ = 1

, i.e., when

θ = 45°

. This only applies when air resistance is negligible and the projectile lands at the same level.

Worked Example03

A ball is launched at 25 m s⁻¹ at an angle of 30° to the horizontal. Calculate the maximum height reached.

1

Find vertical component of initial velocity:

u_{y} = u sin θ = 25 \times sin 30° = 25 \times 0.5 = 12.5 m s^{- 1}

2

At maximum height, $v_{y} = 0$ :

Use

v_{y}^{2} = u_{y}^{2} - 2 g H

(taking upward as positive)

0 = 12. 5^{2} - 2 \times 9.81 \times H

H = \frac{156.25}{19.62} = 7.96 m

Answer: Maximum height H = 7.96 m ≈ 8.0 m

Worked Example04

A football is kicked at 20 m s⁻¹ at 40° to the horizontal from level ground. Calculate: (a) the time of flight, (b) the range.

1

Resolve initial velocity:

u_{x} = 20 cos 40° = 15.32 m s^{- 1}

u_{y} = 20 sin 40° = 12.86 m s^{- 1}

2

(a) Time of flight: For level ground,

T = \frac{2 u _{y}}{g}

T = \frac{2 \times 12.86}{9.81} = 2.62 s

3

(b) Range:

R = u_{x} \times T

R = 15.32 \times 2.62 = 40.1 m

Answers: (a) T = 2.62 s, (b) R = 40.1 m

Worked Example05

A projectile must travel a horizontal distance of 50 m. If launched at 25 m s⁻¹, find the two possible launch angles (assuming level ground).

1

Use range formula:

R = \frac{u ^{2} s i n 2 θ}{g}

50 = \frac{2 5 ^{2} s i n 2 θ}{9.81}

2

Solve for sin 2θ:

sin 2 θ = \frac{50 \times 9.81}{625} = 0.7848

3

Find 2θ:

2 θ = sin^{- 1} (0.7848) = 51.7°

or

180° - 51.7° = 128.3°

4

Find θ:

θ = 25.9°

or

θ = 64.1°

Answer: θ = 25.9° or θ = 64.1°

Worked Example06

A ball is thrown at 18 m s⁻¹ at 35° above horizontal from ground level. It lands on a platform 3 m above the ground. Find the horizontal distance to the platform.

1

Resolve:

u_{x} = 18 cos 35° = 14.74 m s^{- 1}

,

u_{y} = 18 sin 35° = 10.32 m s^{- 1}

2

Vertical motion:

y = u_{y} t - \frac{1}{2} g t^{2}

with y = 3 m

3 = 10.32 t - 4.905 t^{2}

4.905 t^{2} - 10.32 t + 3 = 0

3

Solve quadratic: Using

t = \frac{10.32 \pm 106.5 - 58.86}{9.81}

t = \frac{10.32 \pm 6.90}{9.81}

t = 1.76 s

(ascending) or

t = 0.35 s

(descending - ball landing on way up)

4

Horizontal distance (using t = 1.76 s for far point):

x = u_{x} t = 14.74 \times 1.76 = 25.9 m

Answer: x = 25.9 m (or 5.2 m if landing on the way up)

Worked Example07

A ball is projected at 25 m s⁻¹ at 50° above horizontal from a platform 20 m above the ground. Calculate: (a) the maximum height above the ground, (b) the time of flight, (c) the speed at which it hits the ground.

1

Resolve:

u_{x} = 25 cos 50° = 16.07 m s^{- 1}

,

u_{y} = 25 sin 50° = 19.15 m s^{- 1}

2

(a) Max height above launch point: At max height,

v_{y} = 0

Using

v_{y}^{2} = u_{y}^{2} - 2 g h

:

0 = 19.1 5^{2} - 2 (9.81) h

h = \frac{366.7}{19.62} = 18.7 m

Max height above ground = 20 + 18.7 = 38.7 m

3

(b) Time of flight: Ball lands at y = -20 m (below launch point)

- 20 = 19.15 t - 4.905 t^{2}

→

4.905 t^{2} - 19.15 t - 20 = 0

t = \frac{19.15 + 366.7 + 392.4}{9.81} = \frac{19.15 + 27.56}{9.81} = 4.76 s

4

(c) Final speed:

v_{x} = 16.07 m s^{- 1}

(unchanged)

v_{y} = u_{y} + g t = 19.15 + 9.81 (4.76) = - 27.56 m s^{- 1}

(or use

v_{y}^{2} = u_{y}^{2} + 2 g \times 20

)

v = 16.0 7^{2} + 27.5 6^{2} = 258.2 + 759.6 = 31.9 m s^{- 1}

Answers: (a) 38.7 m, (b) 4.76 s, (c) 31.9 m s⁻¹

Friction, Drag and Air Resistance

When an object travels through a medium, it interacts with particles in that medium. This can slow the object down or provide lift.

Friction

Resistance to motion due to contact forces

Always opposes motion

Converts kinetic energy to heat

Static friction (limiting friction) - force to start motion

Kinetic friction usually less than static

Factors Affecting Friction

Nature of the two surfaces

Normal reaction between surfaces

Almost independent of relative velocity

Drag

Drag is a frictional force experienced by objects travelling through a fluid (liquid or gas). It is caused by collisions between the object and particles in the medium.

Lift and Drag Forces

Bernoulli Principle

The pressure within a fast-moving fluid is lower than in a similar fluid at rest or moving slowly.

How Lift is Generated

Air moves faster over top of wing

Lower pressure above wing

Pressure difference creates upward force

Also: wing pushes air down → reaction force up (Newton's 3rd Law)

Factors Affecting Drag

Speed - faster = more drag

Surface area - larger = more drag

Shape - streamlined = less drag

Fluid density - denser = more drag

Effect of Air Resistance on Projectiles

In reality, air resistance (drag) affects projectile motion significantly:

Effects of Drag

Reduced maximum height

Shorter range

Steeper descent than ascent

Optimal angle less than 45°

Key Facts

Air resistance increases with speed

Often modelled as

F \propto v

or

F \propto v^{2}

Depends on surface area and shape

Terminal Speed

Definition

Terminal speed (or terminal velocity) is the constant velocity reached when the drag force on a falling object equals its weight, resulting in zero net force and zero acceleration.

Key Relationship

As speed ↑, drag ↑ (D ∝ v²) → resultant force ↓ → acceleration ↓ → eventually D = W → a = 0 → constant velocity (gradient = 0 on v-t graph)

Parachutes and Vehicle Maximum Speed

Skydiver with Parachute

1

Initially accelerates at g (D = 0)

2

Speed ↑ → drag ↑ → acceleration ↓

3

Reaches terminal speed (~55 m/s)

4

Parachute opens → area ↑ massively

5

D > W → decelerates (slows down)

6

New lower terminal speed (~5 m/s)

Vehicle Maximum Speed

A vehicle reaches maximum speed when:

Driving Force = Resistive Forces

To increase max speed: More power or reduce drag/friction

Factors Affecting Terminal Velocity

Mass ↑ → v_t ↑ (needs more drag)

Area ↑ → v_t ↓ (more drag at lower v)

Streamlined → v_t ↑ (less drag)

Worked Example08

Explain why the optimal angle for maximum range of a real projectile (with air resistance) is less than 45°.

1

Air resistance acts in the opposite direction to velocity, reducing both horizontal and vertical components of velocity throughout the flight.

2

A projectile at 45° spends more time in the air than one at a lower angle, because it reaches a greater height.

3

The longer the projectile is in the air, the more its horizontal velocity is reduced by air resistance.

4

A lower launch angle means less time in the air, so the horizontal velocity is reduced less by drag.

5

The optimal angle balances height (time of flight) against maintaining horizontal velocity. With air resistance, this optimum shifts below 45°, typically to around 30-40° depending on the projectile.

Answer: Air resistance reduces horizontal velocity throughout flight; lower angles mean less time in air and less velocity loss, so optimal angle is below 45°.

Practice

Question 1: A tennis ball is hit horizontally at 30 m s⁻¹ from a height of 2.5 m. Calculate the horizontal distance it travels before hitting the ground.

1

Time to fall:

h = \frac{1}{2} g t^{2}

2.5 = \frac{1}{2} \times 9.81 \times t^{2}

t = \frac{5}{9.81} = 0.714 s

2

Horizontal distance:

x = u_{x} t = 30 \times 0.714 = 21.4 m

Answer: 21.4 m

Practice

Question 2: A javelin is thrown at 28 m s⁻¹ at 42° to the horizontal. Calculate: (a) the maximum height, (b) the total time of flight if it lands at the same level.

1

Vertical component:

u_{y} = 28 sin 42° = 18.74 m s^{- 1}

2

(a) Max height: Use

v_{y}^{2} = u_{y}^{2} - 2 g H

with

v_{y} = 0

H = \frac{u _{y}^{2}}{2 g} = \frac{18.7 4 ^{2}}{2 \times 9.81} = 17.9 m

3

(b) Time of flight:

T = \frac{2 u _{y}}{g} = \frac{2 \times 18.74}{9.81} = 3.82 s

Answers: (a) H = 17.9 m, (b) T = 3.82 s

Practice

Question 3: A skydiver with mass 75 kg falls with a parachute of cross-sectional area 28 m². At terminal velocity, the drag force equals weight. If the drag coefficient is 1.3 and air density is 1.2 kg m⁻³, use

F = \frac{1}{2} Cρ A v^{2}

to estimate terminal velocity.

1

At terminal velocity: Drag = Weight

\frac{1}{2} Cρ A v^{2} = m g

2

Rearrange:

v = \frac{2 m g}{Cρ A}

3

Substitute values:

v = \frac{2 \times 75 \times 9.81}{1.3 \times 1.2 \times 28}

v = \frac{1471.5}{43.68} = 33.7 = 5.8 m s^{- 1}

Answer: Terminal velocity ≈ 5.8 m s⁻¹ (about 21 km/h)

Summary

Key Equations

QuantityEquation

Component velocities

u_{x} = u cos θ

,

u_{y} = u sin θ

Horizontal motion

x = u_{x} t

(constant velocity)

Vertical motionSUVAT equations with

a = g

Range (level ground)

R = \frac{u ^{2} s i n 2 θ}{g}

Max height

H = \frac{u ^{2} s i n ^{2} θ}{2 g}

Resultant velocity

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

Direction of motion

tan θ = \frac{v _{y}}{v _{x}}

Exam Tips

Always resolve velocity into components first

Use vertical motion to find time (horizontal has no time-dependent acceleration)

State your sign convention clearly

For impact velocity, find both components then use Pythagoras

Remember: air resistance reduces range and makes descent steeper