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Projectile Range

This gives the horizontal distance a projectile travels before landing at the same height it was launched from. It equals velocity squared times the sine of twice the launch angle, divided by gravity. The maximum range occurs at 45 degrees. For example, a ball kicked at 20 m/s at 45 degrees travels about 41 metres. Air resistance makes real ranges shorter, but this ideal formula is the starting point in any physics course.

Physics Mechanics Daily Life
VORZEP · PROJECTILE & BALLISTICS TOOLS

Projectile Range Calculator

Solve the projectile-range relation R = v²·sin(2θ)/g for any of its three unknowns — horizontal range, launch speed, or launch angle — with step-by-step math, a live trajectory schematic, and companion ballistics metrics including maximum height, flight time, and apex velocity.

R = v² · sin(2·θ) / g
Solve for
Horizontal Range (R) Horizontal distance from launch to landing, measured on a level surface. Maximum range occurs at θ = 45° and equals v²/g. Typical: 7.3 m for a long jump, 63 m for a football punt, 90 m for a javelin, 141 m for a baseball throw (with drag), 25 km for an artillery shell, 500 km for an ICBM.
m
Range must be greater than zero.
= 40.775 m = 133.78 ft · H = 10.194 m · T = 2.883 s
Launch Speed (v) The magnitude of the initial velocity vector at launch, in metres per second. Range scales with v², so doubling speed quadruples range. Typical: 10 m/s for a thrown ball, 20 m/s for a football punt, 40 m/s for a baseball throw, 70 m/s for a golf drive, 100 m/s for a tennis serve, 300 m/s for a rifle bullet, 1500 m/s for an artillery shell.
m/s
Launch speed must be greater than zero.
= 72.00 km/h = 44.74 mph · v² = 400.0 m²/s²
Launch Angle (θ) The angle between the initial velocity vector and the horizontal, in degrees. Range is maximum at θ = 45° and equals v²/g. Complementary angles θ and 90° − θ give the same range but different heights. At θ = 0° and θ = 90° the range is zero.
°
Launch angle must be between 0° and 90°.
= 0.7854 rad · sin(2θ) = 1.0000 · complementary angle: 45.00°
Gravitational Acceleration g = 9.810 m/s²
m/s²
HORIZONTAL RANGE (R)
40.7747 m
133.78 ft · H = 10.194 m · T = 2.883 s
SPORTS-PROJECTILE RANGE
R = v² · sin(2θ) / g
Step-by-step solution
Range vs. launch angle
v × 0.7 Current v v × 1.4
Range vs. launch speed
θ = 30° Current θ θ = 60°
Sensitivity
Companion ballistic metrics
Projectile trajectory — live range schematic
PROJECTILE TRAJECTORY · R = v² · sin(2θ) / g R = 40.77 m H = 10.19 m v v_x v_y R = v² · sin(2θ) / g v = 20.00 m/s · θ = 45.00° · g = 9.81 m/s²
Launch and landing points Parabolic trajectory Horizontal range R (apex height H shown dashed)
What the projectile-range relation really measures
Rrange
=
v² sin(2θ)speed and angle
÷
ggravity

Forty-five degrees maximises range

Because R depends on sin(2θ), it reaches its peak when 2θ = 90°, i.e. θ = 45°. The maximum range is R_max = v²/g. This is the single most famous result in elementary ballistics, and it is why athletes, gunners, and rocketeers all default to 45° when they want maximum distance on a level field. Deviating in either direction reduces range symmetrically, because sin(2θ) = sin(180° − 2θ) — so a 30° launch and a 60° launch carry the same horizontal distance.

Range scales with the square of launch speed

Doubling the launch speed quadruples the range, because R ∝ v². This is why a 100 m/s punt travels four times as far as a 50 m/s punt, and why a rifle bullet outranges a thrown ball by orders of magnitude. The v² dependence is a direct consequence of the kinetic-energy scaling: the horizontal distance is proportional to the initial kinetic energy per unit mass, and kinetic energy goes as v². In practice, air resistance puts a hard ceiling on this scaling — a real projectile launched at 1000 m/s does not travel four times as far as one at 500 m/s.

Complementary angles share a range

Because sin(2θ) = sin(180° − 2θ), the angles θ and 90° − θ produce identical ranges. A 30° launch and a 60° launch travel exactly the same horizontal distance, but the 60° projectile reaches twice the maximum height and stays aloft twice as long. This symmetry is only exact in the vacuum approximation — with air resistance, the lower angle wins, because the projectile spends less time at high drag. A real rifleman aiming for maximum range uses about 30–35°, not 45°.

Real projectiles fall short

The vacuum formula is an upper bound. Air resistance removes energy continuously, so a real projectile always lands closer than the formula predicts. The deficit grows with the ballistic parameter β = ρ C_d A v² / (mg), which compares aerodynamic drag to gravity. A golf ball (β ≈ 0.3) loses about 40% of its vacuum range; a baseball (β ≈ 0.1) loses about 25%; a javelin (β ≈ 0.05) loses about 10%; an artillery shell (β ≈ 0.5) loses nearly half. The formula remains useful as a first estimate and for heavy projectiles at low speed, but precise artillery tables must be computed with a full drag model.

How to Use This Calculator

The projectile-range relation links three quantities: R, v, and θ, with g as a physical constant. Pick which one you want the tool to solve, then fill in the other two.

1. Choose what to solve for

Pick R, v, or θ from the toggle. The calculator rearranges R = v²·sin(2θ)/g automatically for whichever variable you choose, and checks that a solution exists.

2. Use realistic presets

Range presets span 1 m to 100 km; speed presets span 5 m/s to 500 m/s; angle presets span 15° to 85°. The g row offers quick switches between Earth, Moon, Mars, Jupiter, and the Sun.

3. Watch the two-solution case

When solving for θ with a given R and v, there are always two solutions — complementary angles that both hit the same range. The calculator returns the lower angle (≤ 45°) as the primary answer and shows the complementary angle in the steps and hints.

4. Respect the physical constraint

A solution for θ exists only if R ≤ v²/g. If you enter a range longer than the maximum possible at that speed, the calculator will flag it. The velocity must be positive; the angle must be between 0° and 90° (exclusive, since both extremes give R = 0).

5. Read the result and verdict

The result card shows the solved quantity plus maximum height H, time of flight T, and apex time. A colour-coded verdict classifies the range from hand-throw through sports, artillery, and missile regimes.

6. Check the schematic and charts

The schematic draws the actual parabolic trajectory to scale, with the launch angle, apex marker, range dimension arrow, and height all labelled. The basic chart plots R against θ for three speeds; the advanced chart plots R against v for three angles.

7. Switch to Advanced for depth

Advanced mode adds the R vs. v chart, a sensitivity check on a +1% input change, and companion metrics including maximum height, time of flight, apex time, impact speed, H/R ratio, and the range at 45° and at the complementary angle.

What This Tool Offers

Built around the projectile-range relation R = v²·sin(2θ)/g, the classic formula for how far a ballistic projectile travels on level ground.

Three-way solving

Solve for horizontal range, launch speed, or launch angle. When solving for θ, the calculator returns the lower complementary solution and reports both — and enforces the R ≤ v²/g feasibility condition.

Live trajectory schematic

A to-scale parabolic trajectory with the launch angle arc, velocity vector and its components, apex marker, range dimension arrow, and height indicator — all updated live.

Range-class verdict

Each result is classified from hand-throw through sports-ball, sports-projectile, artillery, long-range, strategic, and intercontinental regimes, with a coloured badge.

Planetary gravity presets

A dedicated g row with buttons for the Moon (1.62), Mars (3.71), Earth (9.81), Jupiter (24.79), and the Sun (274 m/s²), plus a slider for any value in between.

Dual comparison charts

R vs. θ for three launch speeds showing the sin(2θ) symmetric shape with its maximum at 45°, plus an advanced R vs. v chart showing the quadratic dependence on speed.

Reference tables

A trajectory converter, a lookup grid of R against v and θ, a table of real-world projectile ranges, and a full variable reference.

Reference Tables & Conversions

A full trajectory converter, an R lookup grid, and real-world projectile ranges.

Full trajectory converter

m/s
°
m/s²
m
m
s
v = 20.00 m/s · θ = 45.00° · g = 9.81 m/s² → R = 40.775 m · H = 10.194 m · T = 2.883 s

Range Lookup at g = 9.81 m/s²

Computed from R = v²·sin(2θ)/g. Values in metres. Note θ = 90° always gives R = 0.

v (m/s)θ = 15°θ = 30°θ = 45°θ = 60°θ = 75°θ = 90°

Real-World Projectile Ranges

Representative values. Vacuum formula results overstate real ranges because of air resistance.

Scenariov (m/s)θR vacuum (m)R actual (m)Notes
Child's throw845°6.55.5Light projectile, drag matters
Long jump (takeoff)10.520°7.38.95World record; body rotation helps
Shot put (release)1340°16.923.4World record; heavy ball, low drag
Cricket six (boundary)2535°59.8~75Pull shot, boundary is 70–90 m
Football punt2545°63.755Standard NFL punt
Javelin throw3033°83.398.5World record; javelin is aerodynamic
Soccer goal kick3045°91.775Full pitch length
Baseball throw (outfield)4030°141.390Baseball has high drag
Golf drive7012°168.2270Backspin extends range dramatically
Artillery (WWI howitzer)50045°25,485~12,000Air resistance halves the vacuum range
Paris Gun (WWI)1,60050°256,000130,000First to reach the stratosphere
Modern artillery (base bleed)1,00050°100,00060,000Base-bleed reduces base drag
ICBM (sub-orbital)7,00045°4,994,000~10,000,000Exits atmosphere; vacuum formula fails

Variable Reference

SymbolMeaningTypical rangeUnits
RHorizontal range0.1 – 10⁶m
vLaunch speed1 – 10⁴m/s
θLaunch angle0.1 – 89.9degrees
gGravitational acceleration1.62 – 274m/s²
2θDouble angle (the argument of sin)0.2 – 179.8degrees
v_xHorizontal velocity componentv cos θm/s
v_y0Vertical launch velocityv sin θm/s
HMaximum heightv² sin²θ / (2g)m
TTotal time of flight2v sinθ / gs
R_maxMaximum possible range at speed vv² / g (at θ = 45°)m

What Does R = v² · sin(2θ) / g Measure?

The projectile-range relation connects three quantities at the heart of elementary ballistics: the horizontal distance a projectile travels before returning to its launch height (R), the speed at which it is launched (v), and the angle of the launch above the horizontal (θ), with the gravitational acceleration g setting the scale. Together they answer the question that dominates every sport, every artillery calculation, and every rocketry estimate: how far will it go? The formula is a direct consequence of the parabolic trajectory, and it makes concrete two of the most famous facts in mechanics — that 45° maximises range, and that range grows as the square of speed.

Projectile Range Formula

The defining relationship, obtained by setting the vertical displacement to zero at the landing time, is:

R = v² · sin( 2 · θ ) / g

Rearranged to solve for the other two unknowns:

v = √( R · g / sin(2θ) )
θ = ½ · arcsin( R · g / v² )

Note that the angle equation has two solutions — the complementary angles θ and 90° − θ produce the same range. Written in terms of the velocity components v_x = v cos θ and v_y0 = v sin θ, the range becomes:

R = ( 2 · vx · vy0 ) / g

which makes explicit that range is proportional to the product of the horizontal speed and the vertical speed. The maximum possible range for a given launch speed occurs at θ = 45° and equals:

Rmax = v² / g

Projectile Range Variables

SymbolMeaningNotes
RHorizontal rangeDistance from launch to landing on level ground, in metres
vLaunch speedMagnitude of the initial velocity vector, m/s
θLaunch angleAbove the horizontal; the range is maximised at 45°
gGravitational acceleration9.81 m/s² at Earth's surface, varies with location
v_xHorizontal velocity component= v cos θ, unchanged throughout the flight
v_y0Vertical launch velocity= v sin θ, the component that sets height and time
R_maxMaximum range at speed v= v²/g, achieved at θ = 45°

How Each Variable Affects the Range

Range is proportional to the square of the launch speed, so doubling v quadruples R. It depends on the launch angle through sin(2θ), which is symmetric about θ = 45° and peaks at that angle. Range is inversely proportional to gravity, so a projectile on a low-gravity body travels proportionally farther — the same 20 m/s kick travels 40.8 m on Earth but 247 m on the Moon.

The sin(2θ) dependence deserves close attention. At θ = 0° (horizontal launch), sin 0° = 0, so range is zero — the projectile lands immediately. At θ = 90° (vertical launch), sin 180° = 0, so range is again zero — the projectile goes straight up and comes straight back down. At θ = 45°, sin 90° = 1, the maximum. In between, the curve is a smooth arch, symmetric about 45°, rising steeply from 0° and falling steeply to 90°. At θ = 30°, sin 60° ≈ 0.866, so 86.6% of the maximum; at θ = 60°, sin 120° = sin 60° ≈ 0.866, the same value. This is the geometric reason the two complementary angles produce identical ranges.

The angle at which the range is maximum is only 45° in a vacuum. In real air, drag pushes the optimum downward, because the projectile experiences drag throughout its flight and a flatter trajectory reduces the total time aloft. For a baseball the optimal angle is about 35°; for a golf ball with backspin it can be as low as 12°; for a heavy artillery shell it is around 40°. The vacuum formula remains the foundation of these calculations, but real ballistics tables always deviate from it.

Projectile Range Calculation Examples

Given: v = 20 m/s, θ = 45°, g = 9.81 m/s² — solve for R
2θ = 90°, sin 90° = 1.0000
v² = 400.0
R = 400.0 × 1.0000 / 9.81 = 40.775 m
Given: v = 20 m/s, θ = 30°, g = 9.81 m/s² — solve for R
2θ = 60°, sin 60° = 0.86603
R = 400.0 × 0.86603 / 9.81 = 35.313 m
Given: R = 40.775 m, θ = 45°, g = 9.81 m/s² — solve for v
v = √(40.775 × 9.81 / 1.0000) = √400.0 = 20.000 m/s
Given: R = 35.313 m, v = 20 m/s, g = 9.81 m/s² — solve for θ
sin(2θ) = 35.313 × 9.81 / 400 = 0.86603
2θ = arcsin(0.86603) = 60.00° → θ = 30.00°
The complementary solution is θ = 90° − 30° = 60.00° — same range, different height

Frequently Asked Questions

Because the range is proportional to sin(2θ), and sin reaches its maximum value of 1 when its argument is 90°. Setting 2θ = 90° gives θ = 45°. Intuitively, the projectile needs both horizontal velocity to travel sideways and vertical velocity to stay aloft long enough to travel. These two requirements trade off against each other — a very shallow launch is fast horizontally but lands almost immediately; a very steep launch stays aloft but barely moves sideways. The optimum is the balanced middle case at 45°, where the launch speed is split equally between horizontal and vertical components and the product v_x · v_y0 is maximised.

Because sin(2θ) = sin(180° − 2θ). Setting 2θ₁ = 180° − 2θ₂ gives θ₂ = 90° − θ₁. So any two complementary angles — like 30° and 60°, or 15° and 75° — produce identical ranges. This symmetry is a mathematical consequence of the shape of the sine curve, but it also has a physical reading: a 30° launch throws the projectile nearly flat, so it moves fast but lands quickly. A 60° launch throws it steeply, so it moves slowly but stays aloft much longer. The two effects cancel exactly in the vacuum approximation. In real air the symmetry breaks, because drag depends on speed and the flatter trajectory loses more energy.

In the vacuum approximation, no. Mass cancels out of the equations entirely: the horizontal and vertical motions are independent of mass, and the trajectory depends only on the launch speed and angle. A 1 g pebble and a 1 kg cannonball launched at the same speed and angle travel exactly the same distance. In the real world with air resistance, mass matters a great deal — a heavier projectile has more momentum for the same speed, so drag decelerates it less over a given distance, and it retains more of its speed throughout the flight. This is why lead shot outranges equal-size plastic pellets and why the same cannon firing a heavier shell achieves a longer range.

Air resistance reduces the range, flattens the optimal launch angle, and breaks the complementary-angle symmetry. The drag force is proportional to v² at typical projectile speeds, so a fast projectile loses energy at a rate that grows rapidly with speed. The result is that the projectile's actual path lies entirely inside the vacuum parabola, landing closer and at a shallower impact angle. The magnitude of the effect is captured by the ballistic parameter β = ρ C_d A v² / (mg), which compares the drag force to gravity. For β ≪ 1 (heavy, slow, dense projectiles like cannonballs at moderate speed) the vacuum formula is accurate to a few percent. For β ≳ 0.1 (light, fast, draggy projectiles like golf balls and shuttlecocks) the vacuum range can be 40% too high.

Because drag pushes it downward. A flatter trajectory spends less time in the air and passes through the fast-moving early portion of the flight more quickly, so it loses less total energy to drag. For a baseball thrown at typical outfielder speeds, the optimum is around 35°. For a golf ball with backspin — which actually generates lift — the optimal launch angle can be as low as 12°. For a heavy artillery shell the optimum is around 40°. But the launch conditions also matter: an athlete cannot always achieve the ideal angle given the constraints of their body and the equipment, and the optimal angle for a kicker or a thrower also depends on the ability to impart speed at different angles.

The formula R = v² sin(2θ)/g assumes the projectile launches from and lands at the same height. If it lands on higher ground, the range is shorter than the formula predicts. If it lands in a valley, the range is longer. The general formula for a projectile launched from height h above the landing surface is R = (v cos θ/g) · (v sin θ + √(v² sin²θ + 2gh)). This reduces to the simple form when h = 0. For a cliff launch, the projectile stays aloft longer, and the maximum range occurs at an angle slightly less than 45° — the effect is small for shallow h but significant for launches high above the landing surface.

For a projectile launched and landing at the same level, the maximum height is H = v² sin²θ/(2g) and the range is R = v² sin(2θ)/g. Their ratio is H/R = sin²θ / (2 sin 2θ) = sin²θ / (4 sin θ cos θ) = tan θ / 4. So H = R · tan θ / 4. At 45° this gives H = R/4, meaning the projectile peaks at one quarter of its range — the apex is fairly close to the landing point. At 30° the apex is only R · 0.144, very flat. At 60° the apex is R · 0.433, noticeably taller. At the extremes — θ → 0° or θ → 90° — the range goes to zero and the ratio becomes undefined or infinite.

Only as a rough first estimate. The formula assumes constant gravity, no air resistance, no lift, and a ballistic trajectory with no thrust after launch. A rocket is under thrust for part of its flight, so its trajectory is not a parabola and the range must be computed by integrating the equations of motion with a time-varying thrust. A spinning ball experiences the Magnus force, which can produce lift (backspin) or downward force (topspin), and the trajectory is no longer a parabola. A golf ball with backspin can travel 50–60% farther than the vacuum formula predicts. For any of these cases the vacuum formula is a useful starting point but the real trajectory must be computed numerically.

Calculating...

Variables

Symbol Quantity Unit
R Horizontal range m
v Initial launch speed m/s
theta Launch angle above horizontal degrees
g Gravitational acceleration 9.81 m/s2

What it means

The range of a projectile is the horizontal distance it travels from launch to impact, assuming it lands at the same height. The formula R = v² sin(2θ)/g is derived from kinematics and assumes no air resistance. The range is maximum when θ = 45°. This equation is used in sports (javelin, shot put), military ballistics, and satellite launch calculations. It is also used in engineering for designing trajectories of projectiles and in physics to study motion under gravity. The formula highlights the influence of initial speed and launch angle. In practice, air resistance reduces the range, and corrections are made using more complex models. Understanding projectile motion is fundamental for analyzing motion in two dimensions.

Worked example

Projectile Range – Two Examples

Real‑World
Scenario: A ball is launched at 20 m/s at 45°. Find its range.
ParameterValue
v20 m/s
θ45°
1R = v²·sin(2θ)/g = 400 × sin(90°)/9.81 = 400 × 1 / 9.81 = 40.8 m
Result 40.8 m ✓ Maximum range
Scenario: A projectile is launched at 30 m/s at 30°. Find its range.
ParameterValue
v30 m/s
θ30°
1R = 900 × sin(60°)/9.81 = 900 × 0.866 / 9.81 = 779.4 / 9.81 = 79.4 m
Result 79.4 m ✓ Further
Key insight: Range = v²·sin(2θ)/g – maximum at 45° launch angle.

Common mistakes

  • Angle θ: The launch angle with respect to the horizontal – not the vertical.
  • Initial speed v: The magnitude of the launch velocity – not a component.
  • Range R: Horizontal displacement – only valid if landing height equals launch height.
  • Units: v in m/s, θ in radians/degrees (sine function uses degrees if calculator set), g in m/s² → R in m.
  • Air resistance: This formula neglects air resistance; in reality, range is reduced.

Applications

The range of a projectile, R = v² sin(2θ)/g, gives the horizontal distance travelled by a projectile launched at speed v and angle θ. This formula is essential in ballistics, sports, and military engineering. It is used to calculate the optimal launch angle (45° in vacuum) for maximum range, and to design artillery trajectories and golf drives. In civil engineering, it is applied to the design of water jets and sprinklers. In aerospace, it helps with re‑entry trajectories and landing site prediction. By using this formula, engineers can predict where a projectile will land, which is crucial for safety, accuracy, and efficiency in many practical applications.

  • Ballistic trajectory design for artillery and missiles
  • Sports performance analysis (golf, javelin, long jump)
  • Design of irrigation sprinklers and water jets
  • Re‑entry trajectory prediction for spacecraft
  • Launch angle optimisation for maximum distance