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Projectile motion calculator

Enter the launch speed, angle and height to get the time of flight, range, maximum height and landing speed of a projectile. You can also find the angles that hit a target. The trajectory is drawn and the working is shown.

Units: metres, seconds, m/s and m/s². Free to use, with no sign-up, no adverts and no personal tracking. A learning and checking tool: check the signs and your exam board’s value of g.

Start from an example What do you want to find?

Launch speed u (m/s)Launch angle above horizontal (°)Launch height h (m)Landing height (m)Position at time t (s), optional

Gravity g g (m/s²) Significant figures

Up is positive. The launch height and landing height are measured from the same reference level, usually the ground. Air resistance is ignored.

Result

2.62 sTime of flight40.2 mHorizontal range8.42 mMaximum height1.31 sTime to highest point15.3 m/sHorizontal velocity12.9 m/sInitial vertical velocity20.0 m/sSpeed on landing40.0° belowLanding angle to the horizontal

With no air resistance the horizontal velocity stays constant and the vertical velocity changes by g every second. The path is a parabola.

Working

Resolve the launch velocity
Horizontal: u_x = u cos θ = 20 × cos 40° = 15.3 m/s
Vertical: u_y = u sin θ = 20 × sin 40° = 12.9 m/s (up is positive).
With no air resistance, u_x stays constant and the vertical acceleration is −g = −9.81 m/s².
Highest point
At the top v_y = 0, so t = u_y / g = 12.9 / 9.81 = 1.31 s.
Height gained = u_y² / 2g = 12.9² / (2 × 9.81) = 8.42 m, so the maximum height is 0 + 8.42 = 8.42 m above the reference level.
Time of flight
Vertical position: y = h + u_yt − ½gt². Set y = 0 m and h = 0 m, so ½gt² − u_yt − (h − y) = 0.
t = [u_y + √(u_y² + 2g(h − y))] / g = [12.8558 + √(12.9² + 2 × 9.81 × 0)] / 9.81 = 2.62 s.
The other root is negative, so it is before the launch and is rejected.
Horizontal range
Range = u_x × t = 15.3 × 2.62 = 40.2 m.
Velocity on landing
v_y = u_y − gt = 12.8558 − 9.81 × 2.62 = −12.9 m/s.
Speed = √(u_x² + v_y²) = √(15.3² + −12.9²) = 20.0 m/s, at 40.0° below the horizontal.

Practise projectile questions

Press the button for a question with random numbers. Work it out on paper first, then type your answer.

How the projectile motion calculator works

A projectile moves under gravity alone, so its horizontal velocity stays constant and its vertical velocity changes at a steady rate. The calculator treats the two directions separately:

Resolve the launch velocity into u cos θ along the ground and u sin θ upward.
Find the time to the top from v_y = 0, and the maximum height from v_y² = u_y² − 2gΔy.
Solve y = h + u_yt − ½gt² for the landing time.
Multiply the landing time by the horizontal velocity to get the range.
Combine the horizontal and vertical velocities for the landing speed and angle.

Sign convention

Up is positive, so the acceleration is −g. The launch angle is measured above the horizontal, and a negative angle means the projectile is thrown downward. The launch height and the landing height are measured from the same level.

Worked example

A ball is kicked horizontally at 15 m/s from the edge of a cliff 45 m above level ground. Ignore air resistance and take g = 9.81 m/s². Find the time of flight, the horizontal range and the speed on landing.

Resolve the launch velocity
Horizontal: u_x = u cos θ = 15 × cos 0° = 15.0 m/s
Vertical: u_y = u sin θ = 15 × sin 0° = 0 m/s (up is positive).
With no air resistance, u_x stays constant and the vertical acceleration is −g = −9.81 m/s².
Highest point
The projectile is launched horizontally, so the highest point is the launch point, 45.0 m.
Time of flight
Vertical position: y = h + u_yt − ½gt². Set y = 0 m and h = 45 m, so ½gt² − u_yt − (h − y) = 0.
t = [u_y + √(u_y² + 2g(h − y))] / g = [0 + √(0² + 2 × 9.81 × 45)] / 9.81 = 3.03 s.
The other root is negative, so it is before the launch and is rejected.
Horizontal range
Range = u_x × t = 15.0 × 3.03 = 45.4 m.
Velocity on landing
v_y = u_y − gt = 0 − 9.81 × 3.03 = −29.7 m/s.
Speed = √(u_x² + v_y²) = √(15.0² + −29.7²) = 33.3 m/s, at 63.2° below the horizontal.

So the ball is in the air for 3.03 s, lands 45.4 m from the cliff and hits the ground at 33.3 m/s, 63.2° below the horizontal.

Range against launch angle at 20 m/s

Launched from level ground at 20 m/s. The range is greatest at 45°, and angles that add up to 90° give the same range.

Angle	Range	Maximum height	Time of flight
15°	20.4 m	1.37 m	1.06 s
30°	35.3 m	5.1 m	2.04 s
45°	40.8 m	10.2 m	2.88 s
60°	35.3 m	15.3 m	3.53 s
75°	20.4 m	19 m	3.94 s

Horizontal launch at 10 m/s from different heights

Launch height	Time of flight	Range
5 m	1.01 s	10.1 m
20 m	2.02 s	20.2 m
45 m	3.03 s	30.3 m

Which formulas does this use?

Projectile range (no drag), Projectile maximum height, Projectile time of flight, Projectile height at a given time, Gravitational field and orbit.

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SUVAT calculator, Equations of motion calculator, Free fall calculator, and the A-level physics equation calculator for the rest of the A-level equations.

Common questions

How do you find the range of a projectile?

Split the launch velocity into horizontal u cos θ and vertical u sin θ parts. Find the time of flight from the vertical motion, then multiply it by the horizontal velocity. On level ground this gives R = u² sin 2θ / g.

What launch angle gives the greatest range?

On level ground the greatest range is at 45°. If the projectile is launched from a height, the best angle is slightly below 45°. Two angles that add up to 90° give the same range from level ground.

How do I find the time of flight?

Solve y = h + u sin θ × t − ½gt² for the time when y equals the landing height. This is a quadratic and the later root is the time of flight. From level ground it is 2u sin θ / g.

How do I find the angle to hit a target?

Choose the target mode, then enter the speed, the horizontal distance and the target height. The calculator solves a quadratic in tan θ. There are usually two angles, a low flat one and a high one, and none if the target is out of reach.

Is the calculator free, and does it keep my numbers?

It is free to use, with no sign-up, no adverts and no personal tracking. The numbers you type stay in your browser.

What do I assume about air resistance and g?

Air resistance is ignored, so these are the idealised answers used in GCSE, A-level and first-year university mechanics. Use g = 9.81 m/s² unless your exam board says 9.8 or 10. You can change g to compare other values or other worlds.