Projectile Range

Equivalent forms

R = \frac{v^2 \sin(2\theta)}{9.8}

R = \frac{2v^2 \sin\theta \cos\theta}{g}

The sin(2θ) factor elegantly encodes the symmetry: complementary angles (e.g., 30° and 60°) give the same range.

Symbol	Name	SI unit	Dimension	Range
$R$	Rangeoutput Horizontal distance traveled by the projectile	m	L	0 – 5000
$v$	Launch Speed Initial speed of the projectile	m/s	LT⁻¹	1 – 200
$θ$	Launch Angle Angle of launch above horizontal (in degrees)	°	dimensionless	0 – 90
$g$	Gravitational Acceleration Acceleration due to gravity at Earth's surface	m/s²	LT⁻²	9.8 – 9.8

Unit systems

Symbol	SI	CGS	Imperial
$R$	m	cm	ft
$v$	m/s	cm/s	ft/s
$θ$	°	°	°
$g$	m/s²	cm/s²	ft/s²

Where it holds

Valid for flat ground, no air resistance, uniform

g = 9.8\,\mathrm{m/s}^{2}

, and non-relativistic launch speeds.

Dimensional analysis

R \to [L]

v^{2}\cdot \sin (2\theta )/g \to [LT^{-1}]^{2}/[LT^{-2}] = [L^{2}T^{-2}]/[LT^{-2}] = [L]

Both sides are

[L] =

meters.

\sin (2\theta )

is dimensionless.

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Discovery

Galileo Galilei · 1638

Derived in Discorsi e dimostrazioni matematiche, where Galileo analyzed parabolic trajectories of cannonballs.

Try this

A soccer player kicks at 25 m/s. At what angle does the ball travel the farthest?

R = v²sin(2θ)/g is maximized when sin(2θ) = 1, i.e., θ = 45°. At 25 m/s: R = 625/9.8 ≈ 63.8 m.

Research status: stable

Real-world applications

Sports trajectory optimization
Artillery and ballistics
Sprinkler coverage design
Long jump technique

Common misconceptions

Air resistance makes optimal angle less than $45^{\circ} in$ practice
Complementary angles $(30^{\circ} and 60^{\circ})$ give equal range but different flight times
The formula assumes flat ground — uphill/downhill changes the optimal angle

Experimental verification

Ballistic range tables, artillery firing data, and high-speed camera trajectory tracking.

Derivation

Split initial velocity into vₓ

= v\cdot \cos \theta and v

ᵧ

= v\cdot \sin \theta

.

Time of flight:

T = 2v\cdot \sin \theta /g

.

Range:

R = v

ₓ

\cdot T = v\cdot \cos \theta \cdot 2v\cdot \sin \theta /g = v^{2}\cdot \sin (2\theta )/g

.

Limiting cases

\theta = 0^{\circ} or 90^{\circ}

⟶

R = 0

Horizontal launch has no hang time; vertical launch has no horizontal component.

\theta = 45^{\circ}

⟶

R = v^{2}/g

(maximum)

\sin (90^{\circ}) = 1

gives the maximum range.

g \to 0

⟶

R \to \infty

Without gravity, the projectile travels in a straight line forever.

What if…

What if on the Moon

(g = 1.62)

?

Range increases by factor $9.8/1.62 = 6.05\times$ . A 63.8 m kick becomes 386 m on the Moon.

What if air resistance is included?

Optimal angle drops below $45^{\circ}$ (typically $30^{\circ}$ – $40^{\circ}$ depending on drag). Range decreases significantly at high speeds.

What if launched from a cliff (height h)?

The formula changes: $R = (v^{2}/2g)[\sin (2\theta ) + \surd (\sin ^{2}(2\theta ) + 8gh/v^{2}\cdot \cos ^{2}\theta )]$ . Higher launch point always increases range.

1

Optimal soccer kick

Given ·

v:: 25
theta:: 45
g:: 9.8

Find · R

Steps

At $45^{\circ}$ : $\sin (2\times 45^{\circ}) = \sin (90^{\circ}) = 1$ (maximum)
$R = v^{2}/g = 625/9.8 = 63.8\,\mathrm{m}$
This is the theoretical maximum range at 25 m/s (no air resistance)

Result ·

R = v^{2}\sin (2\times 45^{\circ})/g = 625\times 1/9.8 = 63.8\,\mathrm{m}

2

Complementary angles give equal range

Given ·

v:: 20
theta 1:: 30
theta 2:: 60
g:: 9.8

Find · R for both angles

Steps

At $30^{\circ}$ : $\sin (2\times 30^{\circ}) = \sin (60^{\circ}) = 0.866$
At $60^{\circ}$ : $\sin (2\times 60^{\circ}) = \sin (120^{\circ}) = 0.866$
$R_{30} = R_{60} = 400 \times 0.866 / 9.8 = 35.3\,\mathrm{m}$
Same range, but $60^{\circ} has$ longer flight time and higher peak

Result · Both give

R = 400\times \sin (60^{\circ})/9.8 = 400\times 0.866/9.8 = 35.3\,\mathrm{m}

Newton's Second Law→Kinetic Energy→Newton's Law of Universal Gravitation→