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Projectile Motion Calculator

Horizontal distance equals initial velocity times cosine of the angle times time

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Horizontal Distance

The horizontal distance traveled by a projectile depends on its initial velocity, launch angle, and time. Since there is no horizontal acceleration (ignoring air resistance), horizontal motion is uniform.

x = v₀ cosθ × t

Vertical Distance

The vertical distance accounts for both the upward component of the initial velocity and the downward pull of gravity. The trajectory reaches its peak when the vertical velocity is zero.

y = v₀ sinθ × t − ½gt²

Horizontal Velocity

The horizontal component of velocity remains constant throughout the flight because gravity acts only vertically. It depends solely on the initial speed and launch angle.

vₓ = v₀ cosθ

Vertical Velocity

The vertical velocity decreases as the projectile rises (gravity decelerates it) and increases as it falls. It equals zero at the apex of the trajectory.

vᵧ = v₀ sinθ − gt

Range (Equal Elevation)

The range formula gives the total horizontal distance when launch and landing elevations are equal. Maximum range occurs at a 45° launch angle in a vacuum.

R = v₀² sin(2θ) / g

How It Works

Projectile motion splits into two independent parts: horizontal (constant velocity, x = v₀cosθ·t) and vertical (constant acceleration due to gravity, y = v₀sinθ·t − ½gt²). Together they trace a parabolic trajectory. This calculator solves for distance, velocity, angle, or time in either direction.

Example Problem

A ball is launched at 20 m/s at 45°. How far does it travel horizontally in 2 seconds, ignoring air resistance?

  1. Identify the knowns: initial velocity v₀ = 20 m/s, launch angle θ = 45°, time t = 2 s.
  2. Choose the unknown and the corresponding equation: horizontal distance x, using x = v₀ cos(θ) × t.
  3. Compute the horizontal velocity component: vₓ = v₀ cos(θ) = 20 × cos(45°) ≈ 14.14 m/s.
  4. Substitute into the distance equation: x = vₓ × t = 14.14 × 2.
  5. Evaluate the multiplication: x ≈ 28.28 m.
  6. State the result: the projectile travels approximately 28.28 m horizontally in 2 seconds — and because horizontal velocity is constant in vacuum, the distance scales linearly with time.

Real projectiles drop slightly below this distance because air resistance subtracts horizontal momentum over the flight; the effect grows with launch speed and surface area.

When to Use Each Variable

  • Solve for Horizontal Distance — when you know the launch velocity, angle, and flight time, e.g., calculating how far a ball travels before landing.
  • Solve for Vertical Distance — when you need the height at a specific time, e.g., determining if a projectile clears a wall at a given range.
  • Solve for Horizontal Velocity — when you need the constant horizontal speed component, e.g., analyzing the forward speed of a launched object.
  • Solve for Vertical Velocity — when you need the instantaneous vertical speed at a given time, e.g., finding the velocity just before impact.
  • Solve for Range — when launch and landing elevations are equal, e.g., calculating total distance for an artillery shell on flat terrain.

Key Concepts

Projectile motion decomposes into two independent components: constant horizontal velocity (no horizontal forces, ignoring air resistance) and uniformly accelerated vertical motion (gravity at 9.81 m/s² downward). The resulting trajectory is a parabola. Maximum range on level ground occurs at 45°, and the time of flight depends only on the vertical component.

Applications

  • Sports science: analyzing basketball shot trajectories, football punt distances, and golf ball flight paths
  • Military ballistics: computing artillery range tables and mortar firing angles for given distances
  • Civil engineering: designing water fountain arcs and drainage outfall trajectories
  • Space exploration: calculating launch angles and velocities for suborbital rocket trajectories

Common Mistakes

  • Forgetting to decompose velocity into horizontal and vertical components — you must use v₀cosθ and v₀sinθ separately, not the full velocity in both equations
  • Using the range formula for unequal launch and landing elevations — R = v₀²sin(2θ)/g only works when the projectile lands at the same height it was launched
  • Ignoring air resistance for high-speed or long-range calculations — drag significantly reduces range and alters the optimal launch angle below 45°

Frequently Asked Questions

How do you calculate projectile motion?

Break the velocity into horizontal and vertical components. Horizontal distance is x = v₀ cos(θ) × t with constant velocity, vertical distance is y = v₀ sin(θ) × t − ½gt² with constant downward acceleration g. Solve whichever pair you need at the matching time t; the trajectory follows a parabolic path on flat ground with no air resistance.

What is the formula for projectile motion?

There are five canonical equations the calculator covers: horizontal distance x = v₀ cos(θ) × t, vertical distance y = v₀ sin(θ) × t − ½gt², horizontal velocity vₓ = v₀ cos(θ), vertical velocity vᵧ = v₀ sin(θ) − g × t, and equal-elevation range R = v₀² sin(2θ) / g. They all assume flat ground, no air resistance, and a constant gravitational acceleration g (about 9.81 m/s² near Earth's surface).

What launch angle gives the maximum range?

On flat ground in a vacuum, 45° maximizes range — sin(2 × 45°) = sin(90°) = 1 is the largest the sine factor can be. With air resistance, the optimal angle drops to roughly 30°–40°, and from an elevated launch point the optimal angle is also less than 45° because the projectile gets extra fall time for free.

How do you find the time of flight for a projectile?

For equal-elevation trajectories, time of flight is t = 2 v₀ sin(θ) / g — the time to reach the apex (v₀ sin(θ) / g) doubled by symmetry. For unequal launch and landing heights, set y to the height difference in the vertical-distance equation y = v₀ sin(θ) × t − ½gt² and solve the resulting quadratic for t.

How does launch height affect projectile range?

A launch above the landing surface increases range because the projectile spends extra time falling beyond the equal-elevation apex. The simple range formula R = v₀² sin(2θ) / g no longer applies — you have to find the actual flight time from the vertical-distance quadratic, then multiply by the horizontal velocity v₀ cos(θ). The optimal launch angle for maximum range also drops below 45° as height increases.

Does air resistance affect projectile motion?

Yes, significantly. Air resistance subtracts horizontal momentum throughout the flight and reduces both the range and the maximum height. The trajectory is no longer a perfect parabola — it droops earlier than the no-drag prediction. These calculator equations assume a vacuum, which works well for short flights and dense, low-speed objects (a tossed ball over 10 m) but breaks down for long-range artillery, light projectiles, or anything at speeds where drag becomes comparable to gravity.

Why does horizontal velocity stay constant?

Gravity acts only vertically. With no horizontal force in a vacuum, Newton's first law says the horizontal component of velocity cannot change — the projectile carries forward at v₀ cos(θ) for the entire flight. Vertical velocity, in contrast, changes uniformly: it slows on the way up, hits zero at the apex, and speeds up symmetrically on the way down.

Reference: Lindeburg, Michael R. 1992. Engineer In Training Reference Manual. Professional Publication, Inc. 8th Edition.

Worked Examples

Sports & Ballistics

How far does a soccer kick travel at 25 m/s and 30°?

A player strikes a ball at 25 m/s with a 30° launch angle on level ground. Use the equal-elevation range equation R = v₀² × sin(2θ) / g to get the no-drag horizontal distance.

  • Known: v₀ = 25 m/s, θ = 30°, g = 9.81 m/s²
  • Formula: R = v₀² × sin(2θ) / g
  • Substitute: R = 25² × sin(60°) / 9.81
  • Evaluate: R = 625 × 0.866 / 9.81

Range ≈ 55.18 m on flat ground

Air resistance shortens real soccer kicks by several meters; treat this as the no-drag upper bound, not a target distance.

Architecture & Clearance

How high is a projectile 1.5 seconds after launching at 30 m/s and 45°?

Use the vertical-distance equation y = v₀ sin(θ) × t − ½gt² when the time of flight is known and you need the height at that instant — typical for fence clearance and trajectory-apex estimates.

  • Known: v₀ = 30 m/s, θ = 45°, t = 1.5 s, g = 9.81 m/s²
  • Formula: y = v₀ × sin(θ) × t − ½ × g × t²
  • Substitute: y = 30 × sin(45°) × 1.5 − ½ × 9.81 × 1.5²
  • Evaluate: y = 31.82 − 11.04

y ≈ 20.78 m above the launch point at t = 1.5 s

This is the height at one specific time, not the maximum height. The apex occurs at t = v₀ sin(θ) / g ≈ 2.16 s for this launch.

Engineering & Time of Flight

How long does a projectile stay airborne at 40 m/s and 60°?

For equal-elevation launch and landing, total flight time reduces to t = 2 × v₀ × sin(θ) / g. Solve the vertical-distance equation for t with the final height y set to zero.

  • Known: v₀ = 40 m/s, θ = 60°, final height y = 0, g = 9.81 m/s²
  • Formula: t = (v₀ sin(θ) ± √(v₀² sin²(θ) − 2g × y)) / g
  • With y = 0: t = 2 × v₀ × sin(θ) / g
  • Substitute: t = 2 × 40 × sin(60°) / 9.81 = 80 × 0.866 / 9.81

Time of flight ≈ 7.06 s

Half this time (≈ 3.53 s) is when the projectile reaches its apex. For unequal launch and landing heights, set y to the actual height difference and solve the full quadratic.

Projectile Motion Formulas

All five equations split the velocity vector into a constant horizontal component and a gravity-affected vertical component:

Horizontal distance equals initial velocity times cosine of the angle times timeVertical distance equals initial velocity times sine of the angle times time minus one half times gravity times time squaredHorizontal velocity equals initial velocity times cosine of the angleVertical velocity equals initial velocity times sine of the angle minus gravity times timeRange equals initial velocity squared times sine of two times the angle, divided by gravity

Where:

  • x, y — horizontal and vertical displacement (m)
  • v₀ — initial launch speed (m/s)
  • θ — launch angle measured from horizontal (degrees)
  • t — elapsed time since launch (s)
  • g — gravitational acceleration, ≈ 9.81 m/s² on Earth's surface
  • vₓ, vᵧ — horizontal and vertical velocity components (m/s)
  • R — total horizontal range at equal launch and landing elevation (m)

All five assume a vacuum (no air resistance) and a constant downward gravity. They work well for short flights, dense slow projectiles, and textbook problems; air drag corrections become important for long-range artillery, light objects, or any speed where the drag force is comparable to gravity.

Projectile Trajectory Diagram

The trajectory is a parabola — symmetric on flat ground, with the apex at the midpoint of the horizontal range. The launch velocity v₀ splits into a constant horizontal component v₀ cos(θ) and a vertical component v₀ sin(θ) that slows under gravity, hits zero at the apex, and reverses on the way down.

v₀θHR

v₀ — initial speed and direction · θ — launch angle from horizontal · H — maximum height at the apex · R — horizontal range at equal launch and landing elevation.

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