Horizontal range
Horizontal distance to the return at launch elevation.
Physics · Mechanics · Motion & Energy
Projectile motion calculator — range, time and height
Find range, flight time and maximum height from launch speed and angle. Launch and landing must be at the same elevation; air resistance is neglected.
Created v1.0
By Thibaut Grzelak, Mechanical Analysis Engineer
02
Results
Flight time
s
Time to return to launch elevation.
Maximum height
Height above the launch point.
Horizontal speed
Constant horizontal component in this ideal model.
Initial vertical speed
Upward component at launch; it decreases under gravity.
Relation and numerical substitution
R = v₀² sin(2θ)/g—
Point projectile, no air resistance or wind; uniform constant gravity. Equal launch and landing elevations, locally flat Earth, no Earth rotation. Positive speed and gravity; launch angle greater than 0° and at most 90°.
01Formulas and symbols
Formulas used
R
R = v₀² sin(2θ)/gRange for equal launch and landing elevations.t
t = 2v₀ sin θ/gFlight time from the initial vertical component.hₘₐₓ
hₘₐₓ = v₀² sin² θ/(2g)Height above launch level at zero vertical velocity.vₓ
vₓ = v₀ cos θHorizontal component remains constant.vᵧ₀
vᵧ₀ = v₀ sin θInitial vertical component; later vᵧ = vᵧ₀ − gt.02Assumptions and limits
Scope of validity
- Point projectile, no air resistance or wind; uniform constant gravity.
- Equal launch and landing elevations, locally flat Earth, no Earth rotation.
- Positive speed and gravity; launch angle greater than 0° and at most 90°.
03Validation example
Reference numerical case
- Given v₀ = 20 m/s, θ = 45° = π/4 rad and g = 9.80665 m/s², find the trajectory.
- vₓ = vᵧ₀ = 20/√2 ≈ 14.1421 m/s; t = 2vᵧ₀/g ≈ 2.88419 s; hₘₐₓ = vᵧ₀²/(2g) ≈ 10.1972 m; R = vₓt ≈ 40.7886 m.
- At two significant digits: R ≈ 41 m, t ≈ 2.9 s, hₘₐₓ ≈ 10 m and both initial components ≈ 14 m/s.
04References
Technical references
- University Physics Volume 1, §4.3 Projectile Motion — OpenStax
- Standard acceleration of gravity — NIST
Conventional standard gravity: 9.80665 m/s²; local gravity may differ.
FAQ
Why does 45° maximize range in this ideal model?
For equal launch and landing elevations without drag, R = v₀² sin(2θ)/g. The sine term is largest when 2θ = 90°, so θ = 45° gives the maximum range for a fixed launch speed.
Can I use this calculator when launch and landing heights differ?
No. The displayed range and flight-time relations assume equal elevations. A different landing height requires solving the vertical position equation with the actual height difference.
05How it works
Gravity changes only the vertical component here. The horizontal component stays constant, so range equals horizontal speed multiplied by flight time.
At fixed speed and equal elevations, 45° maximizes ideal range. Complementary angles such as 30° and 60° give the same range but different heights and flight times.
06Limits of the model
Do not use these flight-time or range equations when the landing elevation differs from the launch elevation. Drag can substantially change a real trajectory.
9.80665 m/s² is conventional standard gravity, not a measurement at your location. The full value is retained internally. A vertical launch has zero horizontal range.
07Common mistakes
Trigonometric functions use radians internally; use the angle selector to convert degrees.
Convert launch speed consistently. A low or high angle is not a substitute for modelling drag or different ground levels.