Engineering Mechanics

Projectile Motion Calculator

Calculate range, time of flight, and maximum height for an angled launch, or solve fall time, horizontal distance, and impact speed for a horizontal launch with negligible air resistance.

Formulas R = v₀² sin(2θ) / g · t = √(2h / g)Last updated Aug 14, 2026

Projectile motion combines horizontal motion at constant velocity with vertical motion under constant gravitational acceleration. This calculator covers two practical idealized cases: an angled launch that lands back at the same elevation, and a horizontal launch from a known height. Choose the launch type that matches your problem, then solve for the missing distance, time, or speed metric using standard projectile-motion equations.

Calculation Bench
Launch Type
Solve for
01

g · Positive magnitude of the gravitational acceleration used in the model.

02

v₀ · Initial speed magnitude at the moment of angled launch.

03

θ · Launch angle measured upward from the horizontal.

Angled-launch mode assumes the projectile lands at the same elevation from which it was launched. Use it for textbook range, time-of-flight, and maximum-height problems without drag.

Solution

Enter the required values to calculate range.

R = v₀² sin(2θ) / g

Formula Sheet

T=2v0sin⁡θgT = \dfrac{2v_0\sin\theta}{g}
R=v02sin⁡(2θ)gR = \dfrac{v_0^2\sin(2\theta)}{g}
H=v02sin⁡2θ2gH = \dfrac{v_0^2\sin^2\theta}{2g}
t=2hgt = \sqrt{\dfrac{2h}{g}}
x=vx2hgx = v_x\sqrt{\dfrac{2h}{g}}
v=vx2+2ghv = \sqrt{v_x^2 + 2gh}
  • gGravity
  • v₀Launch Speed
  • θLaunch Angle
  • TTime of Flight
  • RRange
  • HMaximum Height
  • vₓHorizontal Launch Speed
  • hLaunch Height
  • tFall Time
  • xHorizontal Distance
  • vImpact Speed

Variables & Units

SymbolVariableDescriptionCommon Units
gGravityPositive magnitude of the local gravitational acceleration.m/s², ft/s², g
v₀Launch SpeedMagnitude of the initial velocity for an angled launch.m/s, km/h, mph, ft/s
θLaunch AngleLaunch angle measured above the horizontal.deg, rad
TTime of FlightTotal airborne time for a same-elevation angled launch.s, min
RRangeHorizontal distance traveled by an angled launch that lands at its release elevation.m, ft
HMaximum HeightPeak vertical rise above the release point in the angled-launch mode.m, ft
vₓHorizontal Launch SpeedConstant horizontal speed component for a horizontal launch from height.m/s, km/h, mph, ft/s
hLaunch HeightVertical drop from the release point to the landing level in the horizontal-launch mode.mm, cm, m, km, in, ft
tFall TimeElapsed time between horizontal release and landing.s, min
xHorizontal DistanceHorizontal travel distance from launch to landing in the horizontal-launch mode.m, ft
vImpact SpeedMagnitude of the velocity vector just before impact in the horizontal-launch mode.m/s, mph

How to Use This Calculator

  • 01Choose the launch type first. Use Angled Launch on Level Ground when the projectile lands at the same elevation as release. Use Horizontal Launch from Height when the object leaves horizontally from a ledge, table, or platform.
  • 02Select which output to solve for. The calculator will show only the inputs required for that projectile-motion relationship.
  • 03Enter the known values with their units, then select Calculate to see the answer in base SI units, the active formula, and the substitution.
  • 04This tool assumes ideal projectile motion: no air resistance, constant gravity, and no lift, spin, or propulsion after launch.

How the Formula Works

In ideal projectile motion, horizontal and vertical motions are analyzed separately. For an angled launch on level ground, the horizontal velocity stays constant while the vertical component rises and falls under gravity. That leads to the familiar same-elevation formulas T = 2v₀ sinθ / g, R = v₀² sin(2θ) / g, and H = v₀² sin²θ / (2g). These expressions are valid only when launch and landing occur at the same height and air resistance is neglected.

For a horizontal launch from height h, the vertical motion is the same free-fall problem solved in one dimension, so the fall time is t = √(2h / g). The horizontal distance is then x = vₓ t, which gives x = vₓ √(2h / g) after substitution. The impact-speed magnitude combines the unchanged horizontal speed with the gravity-built vertical speed, producing v = √(vₓ² + 2gh).

Worked Example 01

Time of flight for a 45° launch

Known

  • Launch Speed (v₀): 30 m/s
  • Launch Angle (θ): 45°
  • Gravity (g): 9.81 m/s²

Formula

T = 2 v₀ sin θ / g

Substitution

T = (2 × 30 × sin 45°) / 9.81

Result

T ≈ 4.32 s

A 30 m/s launch at 45° stays airborne for about 4.32 s when it lands back at the release elevation.

Worked Example 02

Range on level ground

Known

  • Launch Speed (v₀): 30 m/s
  • Launch Angle (θ): 45°
  • Gravity (g): 9.81 m/s²

Formula

R = v₀² sin(2θ) / g

Substitution

R = (30² × sin 90°) / 9.81

Result

R ≈ 91.74 m

The same 30 m/s, 45° launch travels about 91.74 m horizontally before landing on level ground.

Worked Example 03

Maximum height for an angled launch

Known

  • Launch Speed (v₀): 30 m/s
  • Launch Angle (θ): 45°
  • Gravity (g): 9.81 m/s²

Formula

H = v₀² sin²θ / (2 g)

Substitution

H = (30² × sin²45°) / (2 × 9.81)

Result

H ≈ 22.94 m

At 45°, half of the launch speed contributes to vertical rise, giving a peak about 22.94 m above the release point.

Worked Example 04

Fall time for a horizontal launch

Known

  • Launch Height (h): 45 m
  • Gravity (g): 9.81 m/s²

Formula

t = √(2 h / g)

Substitution

t = √((2 × 45) / 9.81)

Result

t ≈ 3.03 s

A horizontal release from 45 m stays in the air for just over 3 s before impact.

Worked Example 05

Horizontal distance from a platform launch

Known

  • Horizontal Launch Speed (vₓ): 20 m/s
  • Launch Height (h): 45 m
  • Gravity (g): 9.81 m/s²

Formula

x = vₓ √(2 h / g)

Substitution

x = 20 × √((2 × 45) / 9.81)

Result

x ≈ 60.58 m

With 20 m/s horizontal speed and a 45 m drop, the projectile travels about 60.58 m before landing.

Worked Example 06

Impact speed for a horizontal launch

Known

  • Horizontal Launch Speed (vₓ): 20 m/s
  • Launch Height (h): 45 m
  • Gravity (g): 9.81 m/s²

Formula

v = √(vₓ² + 2 g h)

Substitution

v = √(20² + 2 × 9.81 × 45)

Result

v ≈ 35.82 m/s

The unchanged horizontal speed combines with the gravity-generated vertical speed to produce an impact speed of about 35.82 m/s.

Applications

  • 01Introductory engineering mechanics and physics homework involving idealized trajectories
  • 02Estimating throw distance, flight time, or peak height for conceptual motion studies
  • 03Checking how launch speed, angle, or drop height affect two-dimensional motion under gravity

Assumptions

  • 01Air resistance, lift, spin effects, and propulsion after launch are neglected.
  • 02Gravity is treated as constant over the full trajectory.
  • 03Angled-launch formulas on this page apply only when launch and landing are at the same elevation.

Where This Model Stops

  • 01Not suitable for long-range, high-speed, or high-altitude trajectories where drag, wind, or changing gravity matter.
  • 02Does not solve the general case of an angled launch that starts and ends at different elevations; the second mode covers only the special case of a horizontal launch from a known height.
  • 03Does not model projectile motion with air resistance, bounce, roll, or rotating reference-frame effects.

References

  1. [1]
    4.3 Projectile Motion

    OpenStax University Physics Volume 1

    Derives the standard ideal projectile-motion equations, including range, time of flight, and maximum height.

  2. [2]
    Ch. 5 Key Equations

    OpenStax Physics

    Summarizes the projectile-motion equations for range and maximum height in compact reference form.

  3. [3]
    NIST Guide to the SI, Appendix B.9

    NIST

    Provides standard gravity and degree-to-radian conversion factors used in engineering unit work.

Frequently Asked Questions

Why does the 45° rule not always apply?

The textbook result that maximum range occurs at 45° is valid only for ideal projectile motion with no air resistance and equal launch and landing elevations. Once those conditions change, the best angle can shift.

Why does the horizontal-launch mode not ask for an angle?

A horizontal launch starts with an angle of 0° relative to the horizontal, so the initial vertical velocity is zero. The motion then separates cleanly into constant horizontal speed and vertical free fall.

How is this different from the Free Fall Calculator?

Free Fall Calculator handles one-dimensional vertical motion only. This page adds the horizontal component needed for two-dimensional projectile paths, plus same-elevation range and maximum-height formulas.