Projectile Motion Calculator
Calculate the range, maximum height, time of flight, and impact velocity of a projectile launched at any angle — with unit conversion, a full step-by-step solution, and a labeled trajectory diagram.
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Trajectory Diagram
The full flight path of the projectile, drawn to scale from your real numbers. The green arrow is the launch velocity at your angle, the purple point is the peak of the arc with its exact height and timing, and the amber point is where it lands, with the range and total flight time labeled directly on the chart.
Step-by-Step Solution
Here's exactly how this answer was calculated, one step at a time.
Given: v0 = 25 mps, θ = 45°, h0 = 0 m, g = 9.807 m/s²
Step 1: Write down the known values
Every quantity is first converted to a common base unit (meters, seconds, m/s) so the standard projectile-motion formulas can be applied directly.
v0 = 25 mps = 25 m/s θ = 45° h0 = 0 m = 0 m g = 9.807 m/s²Step 2: Resolve the initial velocity into x and y components
The launch velocity is split into a horizontal part vx (which never changes, since gravity only pulls straight down) and a vertical part vy (which gravity constantly slows, stops, then reverses).
vx = v0·cos(θ) = 25·cos(45°) = 17.678 m/s vy = v0·sin(θ) = 25·sin(45°) = 17.678 m/sStep 3: Find the time to reach maximum height
At the very top of the flight the vertical velocity is momentarily zero, so setting vy − g·t = 0 and solving for t gives the time to the peak.
t_peak = vy / g = 17.678 / 9.807 = 1.803 sStep 4: Find the maximum height
This comes from y(t) = h0 + vy·t − ½g·t², evaluated at t = t_peak. It's the highest point the projectile reaches above the ground.
H = h0 + vy² / (2g) = 0 + (17.678)² / (2 × 9.807) = 15.933 mStep 5: Find the total time of flight
This is the positive root of h0 + vy·t − ½g·t² = 0 — the moment the projectile's height returns to y = 0 (the ground). For a launch from ground level (h0 = 0) this simplifies to T = 2vy/g.
T = [vy + √(vy² + 2·g·h0)] / g = [17.678 + √(312.5 + 2×9.807×0)] / 9.807 = 3.605 sStep 6: Find the horizontal range
Since the horizontal velocity never changes, the horizontal distance covered is simply the constant speed vx multiplied by the total flight time T.
R = vx × T = 17.678 × 3.605 = 63.732 mStep 7: Find the impact speed and angle
The horizontal speed stays vx throughout, but the vertical speed at landing is found from vy − g·T. Combining both components gives the final speed and angle at impact.
vy(impact) = vy − g×T = -17.678 m/s Speed = √(vx² + vy(impact)²) = 25 m/s Angle below horizontal = 45°Step 8: Convert the key results to meters (m)
Max height = 15.933 m Range = 63.732 m
Final results:
Range = 63.732 m, Max height = 15.933 m, Time = 3.605 s
Free Online Projectile Motion Calculator with Steps
This projectile motion calculator instantly finds the range, maximum height, time of flight, and impact speed of any object launched into the air at an angle — a football, a cannonball, a basketball shot, an arrow, or a rocket in a physics problem. Enter the initial velocity, the launch angle above the horizontal, an optional launch height, and gravity (Earth, Moon, Mars, or a custom value), and the calculator instantly resolves the motion into horizontal and vertical components, applies the standard kinematics equations, and shows a complete, copyable step-by-step solution alongside a fully labeled trajectory diagram plotted to scale from your exact numbers.
Whether you're a physics student working through projectile motion for the first time, checking homework before an exam, or an engineer or sports analyst estimating how far or how high something will travel, this tool handles launches from ground level as well as launches from an elevated height (a cliff, a building, or a raised platform), and always shows exactly how each result was calculated — not just the final numbers.
What Is Projectile Motion?
Projectile motion describes the curved path traced by any object that is launched into the air and then moves only under the influence of gravity, with air resistance ignored. Once launched, the object's horizontal velocity stays constant the entire flight, while its vertical velocity is continuously slowed, brought to zero at the peak, and then accelerated back downward by gravity — producing the familiar symmetric, parabolic arc you see in the diagram above.
Because the horizontal and vertical motions are completely independent of each other, projectile motion problems are almost always solved by breaking the initial velocity into an x-component and a y-component, solving each direction separately with ordinary constant-acceleration kinematics, and then recombining the results at the end — exactly the approach this calculator uses and shows step by step.
Projectile Motion Formulas
Given an initial speed v0 launched at angle θ above the horizontal from height h0, with gravitational acceleration g, the key formulas are:
- Horizontal velocity component: vx = v0 · cos(θ) — stays constant throughout the flight
- Vertical velocity component: vy = v0 · sin(θ) — changes continuously due to gravity
- Time to reach maximum height: t_peak = vy ÷ g
- Maximum height: H = h0 + vy² ÷ (2g)
- Time of flight (returns to y = 0): T = [vy + √(vy² + 2·g·h0)] ÷ g (simplifies to T = 2vy ÷ g when h0 = 0)
- Horizontal range: R = vx × T
- Position at any time t: x(t) = vx·t and y(t) = h0 + vy·t − ½·g·t²
How to Solve a Projectile Motion Problem Step by Step
Every projectile motion problem, however it's phrased, can be solved with the same sequence of steps:
- Step 1 — List the knowns: initial speed v0, launch angle θ, launch height h0 (0 if not stated), and gravity g (9.81 m/s² on Earth unless stated otherwise).
- Step 2 — Resolve v0 into horizontal and vertical components using vx = v0cos(θ) and vy = v0sin(θ).
- Step 3 — Use the vertical component to find the time to the peak (t_peak = vy/g) and the maximum height (H = h0 + vy²/2g).
- Step 4 — Find the total time of flight T by solving for when the height returns to zero.
- Step 5 — Multiply the constant horizontal velocity by the total flight time to get the range: R = vx × T.
- Step 6 — If needed, find the impact velocity by combining the unchanged horizontal speed with the vertical speed at landing (vy − g·T).
Why the Trajectory Is a Parabola
Substituting t = x/vx into the vertical position equation y(t) = h0 + vy·t − ½g·t² produces y as a function of x that has an x² term, an x term, and a constant — the defining shape of a parabola. This is why every projectile's path, from a thrown ball to a fired artillery shell, traces the same smooth, symmetric curve, always opening downward because gravity is the only force acting on it after launch (ignoring air resistance).
The diagram on this page plots that exact parabola using your real numbers, so you can see visually how the launch angle and speed determine the shape, height, and width of the arc, with the peak and landing point both marked and labeled with their calculated values.
The Optimal Launch Angle for Maximum Range
When the launch and landing heights are equal (h0 = 0), the range formula simplifies to R = v0²·sin(2θ)/g, and since sin(2θ) reaches its maximum value of 1 when 2θ = 90°, the range is maximized at a launch angle of exactly 45°. Any angle above or below 45° — say 30° or 60° — produces the same range as its complementary angle, because sin(2×30°) = sin(2×60°) = sin(60°). This is a classic and frequently tested result in introductory physics.
When the launch height is above the landing height (h0 > 0), however, the optimal angle for maximum range is always slightly less than 45°, since the projectile has extra time to travel horizontally while it falls that additional height. This calculator handles both cases correctly by using the general time-of-flight formula that accounts for a nonzero launch height.
How to Use This Projectile Motion Calculator
Enter the initial launch speed in your preferred unit (meters/second, kilometers/hour, miles/hour, feet/second, or knots), the launch angle in degrees above the horizontal (0° is straight along the ground, 90° is straight up), and the launch height if the object starts above the landing surface (leave it at 0 for ground-level launches). Choose Earth, Moon, or Mars gravity from the dropdown, or select Custom to enter any gravitational acceleration you like — useful for other planets, or for classroom problems that specify g = 10 m/s² for simplicity.
The calculator instantly shows the range, maximum height, total time of flight, time to reach the peak, velocity components, and impact speed and angle, all backed by a complete step-by-step solution and a trajectory diagram plotted to scale with every key value labeled directly on the chart.
Worked Examples
Example 1 — a football kicked at 25 m/s at 45°: vx = 25cos(45°) = 17.68 m/s, vy = 25sin(45°) = 17.68 m/s. Time of flight T = 2×17.68/9.81 = 3.60 s. Range = 17.68 × 3.60 = 63.7 m. Maximum height = 17.68²/(2×9.81) = 15.9 m.
Example 2 — a ball thrown from a 40 m cliff at 15 m/s at 20°: vx = 15cos(20°) = 14.10 m/s, vy = 15sin(20°) = 5.13 m/s. Because h0 = 40 m is not zero, the time of flight uses the full formula: T = [5.13 + √(5.13² + 2×9.81×40)] / 9.81 ≈ 3.36 s, giving a range of about 47.4 m and a maximum height of about 41.3 m above the ground.
Real-World Applications of Projectile Motion
Projectile motion isn't just a textbook exercise — the same equations show up across sports, engineering, and everyday physics:
- Sports science: analyzing the optimal launch angle for a javelin throw, a basketball shot, a golf drive, or a long jump.
- Ballistics and defense: predicting the range and impact point of a fired projectile.
- Firefighting and irrigation: calculating how far a water jet from a hose or sprinkler will travel at a given angle and pressure.
- Space and aerospace engineering: modeling suborbital trajectories and payload drop points, adjusted for different planetary gravities.
- Civil and mechanical engineering: designing safety barriers, catapults, and other systems that launch or must contain moving objects.
- Video game and animation physics engines: simulating realistic arcs for thrown or launched objects.
Tips for Solving Projectile Motion Word Problems Faster
Always split the problem into horizontal and vertical parts immediately — trying to solve for range or height without first finding vx and vy is the most common source of errors. Remember that horizontal velocity never changes during the flight, since gravity has no horizontal component, so vx at launch equals vx at landing. Watch out for problems that give an angle below the horizontal (a downward launch) or ask for quantities partway through the flight rather than at landing — in both cases, use the general position equations x(t) and y(t) directly rather than the shortcut range/height formulas, which only apply to the full flight. And double-check your angle is in the range 0°–90°, since angles outside that range don't correspond to a physical upward launch.
Frequently Asked Questions
What is the formula for projectile motion range?
For a launch and landing at the same height, Range = v0²·sin(2θ) ÷ g. For a launch from a different height, the range is horizontal velocity (v0cosθ) multiplied by the total time of flight, which must be found from the full quadratic time-of-flight equation.
What angle gives the maximum range in projectile motion?
45° gives the maximum range when the launch and landing heights are equal. If the object is launched from above the landing point, the optimal angle is slightly less than 45°.
How do you find the maximum height of a projectile?
Maximum height = launch height + (vertical velocity)² ÷ (2 × gravity), where vertical velocity = v0 × sin(launch angle).
Does air resistance affect this calculator's results?
No — like nearly all introductory physics projectile motion problems, this calculator assumes no air resistance (a vacuum trajectory). Real-world air resistance would reduce the range and height, especially for light or irregularly shaped objects.
Can this calculator handle a launch from an elevated height, like a cliff or building?
Yes — enter a launch height greater than 0, and the calculator automatically uses the general time-of-flight formula that accounts for the extra fall distance.
How is horizontal velocity different from vertical velocity in projectile motion?
Horizontal velocity stays constant throughout the flight because gravity acts only vertically. Vertical velocity continuously decreases on the way up, reaches zero at the peak, then increases in the downward direction on the way back down.