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Projectile Range

Calculates the horizontal distance a projectile travels before landing, assuming it launches and lands at the same height with no air resistance.

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Projectile Range CalculatorR = v² · sin(2θ) / g

R = v² · sin(2·θ) / g
R = range (m)  ·  v = initial velocity (m/s)  ·  θ = launch angle (°)  ·  g = gravity (m/s²)
⟹ SolveR, v, θ, g
m
m/s
°
m/s²
Please fix the errors above.
Solve for:
Presets:
Range
R: v: θ: g:
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Trajectory Sketch
Projectile path Range (R)
R = v²·sin(2θ)/g  ·  For level ground and neglecting air resistance. Maximum range occurs at θ = 45°.

Interpretation

Projectile range: R = v²·sin(2θ)/g, where v is initial speed, θ is launch angle, g is gravity. Gives horizontal distance covered. Example: v=20 m/s, θ=45° → R = 400×1/9.81 ≈ 40.8 m.

R = v^2 * sin(2*theta) / g
Projectile Range

Variables

SymbolQuantityUnit
RHorizontal rangem
vInitial launch speedm/s
thetaLaunch angle above horizontaldegrees
gGravitational acceleration9.81 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

Frequently Asked Questions

Q01What is the formula for the range of a projectile launched from ground level?
A01

The horizontal range (distance travelled) for a projectile launched with initial speed v at angle θ above the horizontal is R = v²·sin(2θ)/g. This assumes the launch and landing heights are equal, and air resistance is neglected.

Q02What is the common mistake when using this formula?
A02

Applying it when the launch and landing heights are different (e.g., launched from a cliff). In that case, you must use the full trajectory equations.

Q03What angle gives the maximum range?
A03

The maximum range is achieved at θ = 45° (since sin(90°) = 1). For any other angle, the range is smaller.

Q04How does the range change with initial speed?
A04

Range is proportional to v². Doubling the launch speed quadruples the range (ignoring air resistance).

Q05How does the range change with gravitational acceleration?
A05

Range is inversely proportional to g. On the Moon (lower g), the range is larger for the same launch conditions.

Q06What is the range if the launch angle is 30°?
A06

R = v²·sin(60°)/g = v²·(√3/2)/g. This is about 86.6% of the maximum range (v²/g).

Q07Does the range depend on the mass of the projectile?
A07

In the absence of air resistance, the range is independent of mass. With air resistance, heavier projectiles tend to have longer ranges.

Q08How do you find the range for a projectile launched from a height h?
A08

Use the equations: t = (v·sinθ + √(v²sin²θ + 2gh))/g (positive root), then R = v·cosθ · t. The formula is more complex than the flat‑ground case.