Projectile range: formula, calculator and worked example

What the symbols mean

$R$: the horizontal range in metres (it lands at the height it was launched from) (m)
$v$: the launch speed in m/s (m/s)
$θ$: the launch angle above the horizontal, between 0° and 90°
$g$: the acceleration due to gravity, 9.81 m/s² on Earth (m/s²)

When to use it

Use it for a ball kicked or thrown on flat ground, a jet of water, or any launch where air resistance is small and the landing height equals the launch height. The range is greatest at 45°, and two angles that add up to 90° (30° and 60°) give the same range. Solving for θ returns both.

Worked example

A ball is kicked at 20 m/s at 45° to level ground (g = 9.81). How far does it go?

Start from the formula.
$R = \frac{v^{2} \sin (2 θ)}{g}$
Substitute $v$ = $20$ , $θ$ = $45 °$ , $g$ = $9.81$ and work it out.
$R = \frac{20^{2} \sin (2 \cdot 45 °)}{9.81}$
$= \frac{400 \sin (90 °)}{9.81}$
$= \frac{400 \cdot 1}{9.81}$
$= \frac{400}{9.81}$
$\approx 40.7747 m$
This assumes level ground (it lands at launch height) and no air resistance. The most range for this speed is $40.7747 m$ , at 45°.

Answer R ≈ 40.77 m

Why it works

Horizontally the projectile moves at a steady $v \cos θ$ ; vertically it goes up and comes back down in time $T = 2vsin θ/g$ . Range is $vcos θ×T = (2v²sin θcos θ)/g$ , and $2sin θcos θ = sin 2θ$ .

Common mistakes

Calculator in the wrong angle mode: sin(90) in radians is not 1.
Writing $\sin^{2} θ$ or $2sin θ$ for $sin 2θ$ .
Using it when landing height differs from launch height (a ball thrown off a cliff); then solve the motion in two directions separately.
Expecting real balls to match: air resistance shortens long, fast flights considerably.

Calculator

See it move

What the symbols mean

When to use it

Worked example

Why it works

Common mistakes