About the Projectile Motion Calculator
This projectile motion calculator solves the classic "ball thrown at an angle" problem. Give it the launch speed, the launch angle above horizontal and, optionally, the height you launch from, and it returns the horizontal range, the maximum height reached, the total time of flight, the time to the peak and the speed at impact — with a plotted trajectory.
Physics students use it to check textbook answers, coaches and players use it to understand why a 45° kick goes farthest on flat ground, and hobbyists use it for water rockets, catapults and trebuchets. You can switch gravity to the Moon, Mars or Jupiter to see how the same throw changes on another world.
The model is ideal projectile motion: gravity is constant, the ground is flat and air resistance is ignored. Real objects (especially light or fast ones) fall short of these numbers because drag slows them, so treat results as the upper limit.
With the default inputs, the horizontal range is 40.79 m. Change any value above to recalculate instantly.
How to use the projectile motion calculator
- 1Enter the launch speed in metres per second.
- 2Enter the launch angle measured up from the horizontal.
- 3Add the launch height if the object starts above the ground it lands on.
- 4Pick the gravity (Earth by default).
- 5Read the range, maximum height and flight time, and check the trajectory chart.
Formula and method
The launch velocity is split into a horizontal component vx = v·cosθ, which stays constant, and a vertical component vy = v·sinθ, which gravity reduces at g metres per second every second. The height is y = h + vy·t − ½g·t²; setting y = 0 and taking the positive root gives the time of flight.
The range is simply the constant horizontal speed multiplied by the flight time, and the maximum height occurs when the vertical velocity reaches zero at t = vy/g. When the launch height is zero these reduce to the familiar R = v²·sin(2θ)/g and H = v²·sin²θ/(2g). Air resistance, wind and the curvature of the Earth are ignored.
- v
- Launch speed (m/s)
- θ
- Launch angle above horizontal
- h
- Launch height above landing level (m)
- g
- Gravitational acceleration (9.80665 m/s² on Earth)
- R, H, t
- Range, maximum height, time of flight
Worked examples
Ball kicked at 20 m/s and 45°
With vx = vy = 14.14 m/s, the ball is in the air for 2 × 14.14 / 9.807 ≈ 2.88 s. It travels 14.14 × 2.88 ≈ 40.79 m and peaks at 14.14² / (2 × 9.807) ≈ 10.20 m.
Throw from a 10 m cliff at 15 m/s, 30°
vy = 7.5 m/s and vx = 12.99 m/s. Solving 10 + 7.5t − 4.903t² = 0 gives t ≈ 2.38 s, so the stone lands about 31.0 m out. The peak is 10 + 7.5² / 19.61 ≈ 12.87 m and it hits the ground at about 20.5 m/s.
Same 45° throw on the Moon
With one-sixth of Earth’s gravity, R = 20² / 1.62 ≈ 246.9 m — about six times farther — and the flight lasts roughly 17.5 s.
Frequently asked questions
What angle gives the maximum range?+
On level ground with no air resistance, 45° gives the longest range because it balances horizontal speed against time in the air. Launching from a height lowers the optimum angle slightly; with air drag it is usually between 30° and 40°.
What is the formula for the range of a projectile?+
For a launch and landing at the same height, R = v² · sin(2θ) / g. For a launch from height h, compute the flight time t = (v·sinθ + √((v·sinθ)² + 2gh)) / g, then R = v·cosθ · t.
How do you find the maximum height of a projectile?+
The peak happens when the vertical velocity reaches zero. Maximum height above the launch point is (v·sinθ)² / (2g), then add the launch height h if the object started above the ground.
Why do complementary angles give the same range?+
Because sin(2θ) = sin(180° − 2θ), angles like 30° and 60° give identical ranges on flat ground. The steeper throw goes higher and stays in the air longer, but moves more slowly horizontally.
Does mass affect projectile motion?+
Not in the ideal model — all objects accelerate downward at g regardless of mass. In real air, heavier and denser objects are affected less by drag, so they come closer to the ideal range.