Range of Projectile Formula: R = u² sin 2θ / g, Examples

The range of projectile formula is $R = \frac{u^2 \sin 2\theta}{g}$ . It gives the horizontal distance a projectile covers before it comes back down to its launch height, for a launch speed u at an angle θ above the horizontal, with air resistance ignored.

Two results make this formula a favourite of examiners. The range is largest at θ = 45°, where it equals $\frac{u^2}{g}$ . And two angles that add up to 90°, such as 30° and 60°, give exactly the same range. Both drop straight out of the sin 2θ term. US books write the same result with v₀ in place of u.

Quick facts

Formula
$R = \frac{u^2\sin 2\theta}{g}$
Maximum range
$R_{max} = \frac{u^2}{g}$ at θ = 45°
Built from
$R = (u\cos\theta) \times T$
SI unit
metre (m)
Same range
Angles θ and 90° − θ
Assumes
Launch and landing at the same height, no air drag

Deriving the range of projectile formula

Horizontally there is no acceleration, so the ball moves at a steady $u\cos\theta$ . It keeps doing that for the whole time of flight, $T = \frac{2u\sin\theta}{g}$ . Distance is speed times time:

$R = u\cos\theta \times \frac{2u\sin\theta}{g} = \frac{u^2 (2\sin\theta\cos\theta)}{g}$

The double angle identity $2\sin\theta\cos\theta = \sin 2\theta$ turns this into $R = \frac{u^2 \sin 2\theta}{g}$ . The time of flight formula page shows how T is found.

What is the formula of range in projectile motion? Worked examples

Example 1: a ball at 30°

A cricket ball leaves the bat at 20 m/s, angled 30° up from flat ground. Find its range.

$R = \frac{20^2 \times \sin 60^\circ}{9.8} = \frac{400 \times 0.866}{9.8} = \frac{346.4}{9.8} \approx 35.3\ \text{m}$

Example 2: the best angle, speed in km/h

A shot put leaves an athlete's hand at 72 km/h. Ignoring the release height, what is the farthest it can travel?

Dividing by 3.6 gives u = 20 m/s. The best angle is 45°, where sin 90° = 1.

$R_{max} = \frac{20^2}{9.8} = \frac{400}{9.8} \approx 40.8\ \text{m}$

Example 3: finding the launch angle

A hose sprays water at 14 m/s and it lands 10 m away on level ground. At what angle is the hose held?

$\sin 2\theta = \frac{gR}{u^2} = \frac{9.8 \times 10}{196} = 0.5$ , so $2\theta = 30^\circ$ or $150^\circ$ .

The hose is at 15° or at 75°. Both give a 10 m range.

When you solve for the angle, always give both answers unless the question rules one out. The calculator returns the smaller one.

Why 45° wins, and the complementary angle rule

The sine of an angle is largest, equal to 1, at 90°. For sin 2θ to reach 1, θ must be 45°. Below 45° the ball moves fast sideways but lands too soon. Above 45° it stays up longer but drifts sideways too slowly. At 45° the two effects balance.

Since sin(180° − x) = sin x, the angles θ and 90° − θ give the same value of sin 2θ. At 20 m/s, launches at 15° and 75° both land 20.4 m away. The 75° launch climbs much higher and stays up much longer to cover the same ground.

Range formula physics: limits and real throws

The formula needs launch and landing at the same height. For a ball thrown from a cliff or a basketball shot that lands above the release point, find the time of flight from the vertical motion first, then multiply by u cos θ.

Air resistance shortens every real range, and the best angle for a real throw is usually a little below 45°. Class 11, JEE and AP Physics problems ignore drag unless they say otherwise, so use the formula as written in exams.

Mistakes that cost marks

Writing sin²θ or 2 sin θ instead of sin 2θ.
Forgetting to square u. Doubling the launch speed makes the range four times larger.
Radian mode on the calculator.
Using it off a cliff, where launch and landing heights differ.

Range shares its inputs with the peak height formula, and at 45° the peak is exactly one quarter of the range. More projectile and straight line results are in the kinematics formulas hub.

Frequently asked questions

Which formula of range of projectile motion do exams expect?

The formula for range of projectile motion on level ground is R = u² sin 2θ / g. It holds when the landing height equals the launch height and drag is neglected.

At what angle is the range of a projectile maximum?

At 45°, because sin 2θ reaches its largest value of 1 there. The maximum range is u² / g.

Why do launches at 30° and 60° land equally far away?

Because sin 60° and sin 120° are equal. Any two launch angles that add to 90° give the same range at the same speed.

What is the unit of range in projectile motion?

Range is a horizontal distance, so it is measured in metres (m).

How does range depend on launch speed?

Range is proportional to u². Doubling the launch speed gives four times the range at the same angle.