Projectile Motion Calculator

The projectile motion calculator solves the complete kinematics of an object launched into the air with an initial speed v₀ at an angle θ above the horizontal, from an optional initial height h₀. It computes the time of flight, the maximum height, the horizontal range, and samples the full parabolic trajectory, all from the same three simple equations that have described falling and thrown objects since Galileo first separated horizontal and vertical motion in the early 1600s.

Projectile motion is one of the cleanest examples in all of physics of a complex-looking problem that decomposes into two much simpler ones. Because gravity acts purely vertically and nothing else acts on the object once it leaves the launcher, the horizontal motion is uniform (constant velocity) and the vertical motion is uniformly accelerated (constant deceleration going up, constant acceleration coming down). Treat the two directions independently, recombine them, and the result is always a parabola.

How to Use the Projectile Motion Calculator

1. Enter the initial velocity v₀ in metres per second, the speed of the object the instant it leaves the launcher, thrower's hand, or barrel.
2. Enter the launch angle θ in degrees, measured from the horizontal (0° is dead flat, 90° is straight up). Valid range is 0°-90°.
3. Optionally enter an initial height h₀ in metres if the object is launched from an high position, a cliff, a building, a table. Leave it at 0 for a flat, ground-level launch.
4. Optionally adjust the gravitational acceleration g (default 9.81 m/s² for Earth), or click one of the planet/moon presets (Moon, Mars, Venus, Jupiter) to see how the same launch behaves elsewhere.
5. Click Calculate. The result panel shows the horizontal range, time of flight, maximum height above the ground, time to reach the apex, the constant horizontal velocity, the vertical velocity at launch, the extra height gained above the launch point, and a 10-point table sampling the (t, x, y) trajectory.
6. For a quick sanity check, click any of the five reference-example chips, including the textbook v₀ = 20 m/s at 45° case, which should return a range of about 40.77 m, a time of flight of about 2.88 s, and a maximum height of about 10.19 m.

The Kinematic Equations

Split the initial velocity into horizontal and vertical components using the launch angle:

vx = v₀ · cos θ (constant for the whole flight, nothing accelerates the projectile horizontally)

vy = v₀ · sin θ (the initial vertical speed, which shrinks under gravity on the way up and grows on the way down)

At any time t after launch, the position is given by the standard constant-acceleration equations:

x(t) = vx · t

y(t) = h₀ + vy · t − ½ g t²

Setting y(t) = 0 and solving the resulting quadratic in t for the positive root gives the general time of flight:

t = [ vy + √(vy² + 2 g h₀) ] / g

When the launch height h₀ is zero, the square root simplifies to vy, and the formula collapses to the familiar t = 2 vy / g. The time to reach the apex (the highest point) is exactly half of this flat-ground time: t_apex = vy / g, because the vertical velocity decreases linearly from vy to 0 on the way up.

The maximum height above the launch point occurs at t_apex, and substituting back gives:

Δh = vy² / (2 g)

The maximum height above the ground is simply h₀ + Δh. Finally, the horizontal range is the horizontal velocity multiplied by the full time of flight:

R = vx · t

common flat-ground case (h₀ = 0), substituting the time-of-flight formula and using the double-angle identity sin(2θ) = 2 sin θ cos θ gives the compact and famous form:

R = v₀² · sin(2θ) / g

History and Context

The systematic study of projectile motion begins with Galileo Galilei, who in Dialogues Concerning Two New Sciences (1638) demonstrated, through careful experiments rolling balls down inclined planes and off tables, that horizontal and vertical motions are independent and that the resulting path is a parabola. This was a radical break from the Aristotelian view, which held that a thrown object moves in a straight line under an "impressed force" until that force is exhausted, after which it falls straight down. Galileo's insight that gravity acts continuously and independently of horizontal motion is the conceptual foundation of everything in this calculator.

The practical motivation was military. Renaissance mathematicians and gunners needed to predict where a cannonball would land, and early ballistics tables (going back to Niccolò Tartaglia in the 1530s) were built from empirical trial and error rather than theory. Galileo's parabolic trajectory model, later refined by Isaac Newton's laws of motion and universal gravitation, turned artillery aiming from an art into a calculable science. Modern exterior ballistics adds corrections for air resistance, wind, the Coriolis effect, and even the curvature and rotation of the Earth for very long-range projectiles, corrections that this calculator, like most introductory treatments, deliberately leaves out to keep the core physics isolated.

Worked Examples

Example 1, The textbook case. A ball is thrown at v₀ = 20 m/s at θ = 45° from flat ground (h₀ = 0), g = 9.81 m/s². The components are vx = 20 cos 45° ≈ 14.142 m/s and vy = 20 sin 45° ≈ 14.142 m/s. Time of flight t = 2 × 14.142 / 9.81 ≈ 2.883 s. Maximum height Δh = 14.142² / (2 × 9.81) ≈ 10.194 m. Range R = 14.142 × 2.883 ≈ 40.77 m (equivalently, R = 20² × sin 90° / 9.81 = 400 / 9.81 ≈ 40.77 m). This is the single most commonly cited example in introductory mechanics, precisely because 45° is the angle that maximises range on flat ground.

Example 2, A shallower launch. A javelin is thrown at v₀ = 30 m/s at θ = 30°, flat ground. vx = 30 cos 30° ≈ 25.98 m/s, vy = 30 sin 30° = 15 m/s. Time of flight t = 2 × 15 / 9.81 ≈ 3.058 s. Range R = 25.98 × 3.058 ≈ 79.45 m. Notice this is nearly twice the range of Example 1 despite a similar-looking angle change, because range scales with v₀², a 50% increase in speed (20 → 30 m/s) increases range by a factor of 2.25 before the angle is even considered.

Example 3, Launched from height. A stone is thrown at v₀ = 15 m/s at θ = 60° from the top of a 10 m cliff. vx = 15 cos 60° = 7.5 m/s, vy = 15 sin 60° ≈ 12.99 m/s. The discriminant is vy² + 2 g h₀ = 168.75 + 196.2 = 364.95, so t = (12.99 + √364.95) / 9.81 ≈ (12.99 + 19.104) / 9.81 ≈ 3.272 s. Range R = 7.5 × 3.272 ≈ 24.54 m. Maximum height above the ground is 10 + 12.99² / 19.62 ≈ 10 + 8.60 ≈ 18.60 m. Compare this to the same launch from flat ground (h₀ = 0): the time of flight would only be 2 × 12.99 / 9.81 ≈ 2.648 s and the range only about 19.86 m, the extra 10 m of launch height adds nearly 24% to the range.

Example 4, Straight up. A ball is thrown straight up (θ = 90°) at v₀ = 50 m/s from flat ground. Here vx = 0, so the horizontal range is exactly zero regardless of the time of flight. vy = 50 m/s, so t = 2 × 50 / 9.81 ≈ 10.19 s and the maximum height is 50² / 19.62 ≈ 127.4 m. This example is a useful edge case for verifying that a calculator correctly reduces to pure vertical (free-fall) motion at the θ = 90° boundary.

Example 5, On the Moon. The same v₀ = 25 m/s at θ = 40° launch, but with g = 1.62 m/s² (the Moon's surface gravity, about 1/6th of Earth's). vx = 25 cos 40° ≈ 19.15 m/s, vy = 25 sin 40° ≈ 16.07 m/s. Time of flight t = 2 × 16.07 / 1.62 ≈ 19.84 s, nearly seven times longer than the same launch on Earth. Range R = 19.15 × 19.84 ≈ 379.9 m, roughly six times the Earth range for the identical launch speed and angle. This dramatic difference is exactly what Apollo 14 astronaut Alan Shepard demonstrated in February 1971 when he hit golf balls on the lunar surface, even a mediocre golf swing sent the ball extraordinarily far, purely because of the Moon's weak gravity, since there is also no atmosphere to add drag.

Common Misconceptions

1. "The projectile speeds up on the way down because gravity is pulling it." Gravity does accelerate the object downward throughout the flight (both going up and coming down), but the vertical speed only exceeds the launch vertical speed once the projectile falls back below the launch height (or, for a flat launch, only at the very instant it returns to the ground, where its vertical speed exactly equals its initial vertical speed but in the opposite direction). The horizontal speed never changes.
2. "A heavier object falls, and is thrown, differently from a lighter one." In the absence of air resistance, all objects experience the same gravitational acceleration g regardless of mass, so trajectory shape depends only on v₀, θ, h₀, and g, not on the mass of the projectile. Mass only matters once air resistance is included, because drag force depends on cross-sectional area and shape, not directly on mass, so lighter or less dense objects (a feather, a badminton shuttlecock) are affected far more than dense compact ones (a shot put).
3. "45° always gives the maximum range." This is only true for a flat-ground launch (h₀ = 0) with no air resistance. When launching from a height, or when air resistance is significant, the optimal angle shifts below 45°. For a launch from an high position, the optimal angle can be found by differentiating the general range formula and is always less than 45°, approaching 0° as the launch height becomes very large relative to the achievable height gain.
4. "The trajectory is symmetric even when launched from a height." On flat ground, the trajectory is a symmetric parabola: the time to rise equals the time to fall, and the impact speed equals the launch speed. When h₀ ≠ 0, this symmetry is broken, the projectile spends more time falling than rising (or vice versa if h₀ is negative, i.e. thrown from below the landing point), and the impact speed differs from the launch speed.
5. "Doubling the speed doubles the range." Because range depends on v₀² (in the flat-ground formula R = v₀² sin 2θ / g), doubling the launch speed quadruples the range, not doubles it, a frequently underestimated relationship in sports and ballistics.

Frequently Asked Questions

Why does the horizontal velocity stay constant? Because, in the idealised no-air-resistance model, no horizontal force acts on the projectile after launch. Gravity acts only vertically (straight down), so by Newton's second law there is no horizontal acceleration, and vx = v₀ cos θ remains constant for the entire flight. This is the key simplification that makes the horizontal and vertical motions independently solvable.

What launch angle gives the maximum range? On flat ground with no air resistance, 45° maximises the range, since R = v₀² sin(2θ) / g is maximised when sin(2θ) = 1, which happens at 2θ = 90°, i.e. θ = 45°. With air resistance, or when launching from or to different heights, the optimal angle shifts, typically lower than 45° for high-speed projectiles moving through air, and lower still when launched from an high position.

Does air resistance change these formulas? Yes, significantly for light or fast-moving objects such as a table-tennis ball, a bullet, or a badminton shuttlecock. Air resistance is usually modelled as a drag force proportional to the square of velocity, which makes the equations of motion nonlinear and generally requires numerical integration rather than the closed-form equations used here. For dense, slow, compact objects over short ranges, a shot put, a cannonball at modest speed, the no-drag approximation used by this calculator is usually accurate to within a few percent.

How do I find the range for a projectile launched from a height, like off a cliff? Use the general time-of-flight formula t = [vy + √(vy² + 2 g h₀)] / g, then multiply by the constant horizontal velocity vx to get the range. The extra √(2 g h₀) term inside the square root accounts for the additional time gained by starting above the landing height. A projectile launched from a cliff always travels farther than the identical launch from flat ground, because it spends more time in the air before it lands.

What happens at θ = 90°, i.e. straight up? At θ = 90°, cos θ = 0, so the horizontal velocity vx = 0 and the horizontal range is exactly zero, the projectile goes straight up and comes straight back down at the same spot (on flat ground). The time of flight and maximum height are still governed by the same vertical equations, with the full initial speed going into the vertical component, vy = v₀.

Why is 9.81 m/s² used for gravity, and does it change by location? 9.81 m/s² is the standard value for Earth's gravitational acceleration at sea level and mid-latitudes. The true value varies slightly with latitude and altitude, from about 9.78 m/s² near the equator (partly due to centrifugal effects of Earth's rotation) to about 9.83 m/s² near the poles, and decreases gradually with altitude. For virtually all everyday and classroom calculations, 9.81 m/s² (or the rounder 9.8 m/s²) is accurate enough.

Can this calculator be used for motion on other planets or the Moon? Yes, simply change the gravitational acceleration g to the value for that body. Presets are provided for the Moon (1.62 m/s²), Mars (3.71 m/s²), Venus (8.87 m/s²), and Jupiter (24.79 m/s²). Since the underlying kinematic equations derive purely from Newton's laws and do not depend on any Earth-specific physics, they apply unchanged to any environment where gravity is effectively the only force acting on the projectile, that is, anywhere with a negligible atmosphere, or in a vacuum.

Why is the maximum-range angle always exactly half the launch-and-return angle range? This follows from the identity sin(2θ) = sin(180° − 2θ). Because the range formula depends only on sin(2θ), any two launch angles θ₁ and θ₂ that satisfy θ₁ + θ₂ = 90° give identical ranges on flat ground (e.g. 30° and 60°, or 15° and 75°). The complementary pair with the steeper angle produces a higher and slower trajectory with a longer time of flight, while the shallower angle produces a flatter, faster trajectory that covers the same distance more quickly.

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