Physics · Classical Mechanics · Kinematics
Projectile Motion Calculator
Calculates the range, maximum height, time of flight, and final velocity of a projectile launched at a given speed and angle.
Calculator
Formula
R is the horizontal range (m), H is the maximum height (m), T is the total time of flight (s), v₀ is the initial launch speed (m/s), θ is the launch angle above horizontal (degrees), and g is the acceleration due to gravity (9.81 m/s²). These equations assume launch and landing at the same elevation, negligible air resistance, and constant gravitational acceleration.
Source: Halliday, Resnick & Krane — Physics, 5th Edition, Chapter 4: Motion in Two and Three Dimensions.
How it works
Projectile motion describes the two-dimensional trajectory of an object launched with an initial velocity at an angle to the horizontal, subject only to gravitational acceleration. Because gravity acts exclusively in the vertical direction, the horizontal and vertical components of motion are mathematically independent. The horizontal component moves at constant velocity throughout the flight, while the vertical component follows uniformly accelerated (decelerated upward, accelerated downward) motion. This separation is the core insight that makes the kinematic equations tractable.
The governing equations are derived from Newton's second law applied separately to each axis. The horizontal range is given by R = v₀² sin(2θ) / g, which shows that maximum range occurs at a 45° launch angle. The maximum height reached is H = v₀² sin²(θ) / (2g), and the total time of flight is T = 2v₀ sin(θ) / g. These three results follow directly from integrating the constant-acceleration equations twice. The horizontal velocity component v_x = v₀ cos(θ) remains constant throughout flight, while the vertical component varies linearly with time. At landing, because the trajectory is symmetric about the apex (for flat ground), the final speed equals the initial speed v₀ — only the direction changes.
Practical applications span an enormous range of scales and disciplines. In sports science, coaches use these equations to optimise throwing or kicking angles for maximum distance. Civil and military engineers apply ballistics analysis to predict impact zones. In physics education, projectile motion is typically the first multi-dimensional problem students encounter, establishing intuition about vector decomposition. Even in video game development, accurate projectile physics relies on these same kinematic relationships to produce realistic trajectories for bullets, grenades, and thrown objects.
Worked example
Suppose a ball is kicked from flat ground with an initial speed of 25 m/s at a launch angle of 40° under standard gravity (g = 9.81 m/s²).
Step 1 — Convert angle to radians: θ = 40° × π/180 = 0.6981 rad.
Step 2 — Horizontal Range: R = (25² × sin(80°)) / 9.81 = (625 × 0.9848) / 9.81 = 615.5 / 9.81 ≈ 62.74 m.
Step 3 — Maximum Height: H = (25² × sin²(40°)) / (2 × 9.81) = (625 × 0.4132) / 19.62 = 258.25 / 19.62 ≈ 13.16 m.
Step 4 — Time of Flight: T = (2 × 25 × sin(40°)) / 9.81 = (50 × 0.6428) / 9.81 = 32.14 / 9.81 ≈ 3.276 s.
Step 5 — Horizontal Velocity: v_x = 25 × cos(40°) = 25 × 0.7660 ≈ 19.15 m/s (constant throughout flight).
Step 6 — Final Speed at Landing: Because air resistance is neglected and the ground is flat, the final speed equals the initial speed: 25.00 m/s (direction is 40° below horizontal by symmetry).
This result confirms that a 40° launch angle yields only a slightly shorter range than the theoretical optimum of 45°, while producing a more realistic low-arc trajectory suitable for many ball sports.
Limitations & notes
This calculator assumes ideal conditions that rarely hold perfectly in the real world. The most significant simplification is the neglect of air resistance (drag). For low-speed, dense objects such as shotputs, the ideal equations are quite accurate, but for lightweight objects (shuttlecocks, ping-pong balls) or high-speed projectiles (bullets, golf balls), aerodynamic drag substantially reduces range and alters the trajectory shape, making it non-parabolic. Similarly, the Magnus effect — spin-induced lift — causes significant deviation in rotating balls. The equations also assume flat, level ground; launching from a cliff or into a valley requires modified formulas accounting for the height difference between launch and landing points. Wind and variations in air density with altitude are ignored. For very long ranges (artillery shells, intercontinental ballistics), the Coriolis effect from Earth's rotation and the curvature of the Earth become non-negligible. Finally, the gravitational acceleration g is treated as constant at 9.81 m/s²; for precise work near different planetary bodies or at high altitudes, the local value of g should be substituted. Despite these limitations, the ideal projectile model remains the essential first approximation and educational foundation for more complex ballistic analyses.
Frequently asked questions
At what angle does a projectile travel the maximum range?
For a projectile launched and landing at the same elevation with no air resistance, the maximum horizontal range is achieved at a launch angle of exactly 45°. This follows directly from the range formula R = v₀² sin(2θ) / g, which is maximised when sin(2θ) = 1, i.e., 2θ = 90°. In real-world conditions with air drag, the optimal angle is typically lower than 45°.
Why does the final speed equal the initial launch speed?
In the ideal case (no air resistance, flat ground), the trajectory is perfectly symmetric about the highest point. The projectile loses kinetic energy to potential energy on the way up and regains it exactly on the way down. By energy conservation, the speed at landing equals the speed at launch — only the direction of the velocity vector has changed (reflected about the horizontal).
Can I use this calculator for projectiles launched on other planets?
Yes. The gravitational acceleration input field is editable, so you can substitute any value of g. For the Moon, use g ≈ 1.62 m/s²; for Mars, use g ≈ 3.72 m/s²; for Jupiter, use g ≈ 24.79 m/s². The equations themselves are universal — only the gravitational constant changes between bodies, assuming the atmosphere is negligible.
How do I account for a launch height above the landing point?
This calculator assumes the launch and landing elevations are identical. If the projectile is launched from a height h above the landing point, the time of flight is found by solving the quadratic y(t) = h + v₀ sin(θ)·t − ½g·t² = 0. The range is then simply v₀ cos(θ) × T. These calculations require a different, more general form of the kinematic equations not covered by the standard flat-ground formulas here.
What is the relationship between the two complementary launch angles that give the same range?
Two launch angles that are symmetric about 45° — for example 30° and 60°, or 20° and 70° — produce identical horizontal ranges (given the same launch speed and flat ground). This is because sin(2θ) = sin(180° − 2θ), so the range formula yields the same result for θ and (90° − θ). However, the higher angle produces a greater maximum height and a longer time of flight, while the lower angle produces a flatter, faster trajectory.
Last updated: 2025-01-15 · Formula verified against primary sources.