Projectile Motion Calculator: Range, Maximum Height and Time of Flight

Launch speed and launch angle in; range, maximum height, time of flight, both velocity components and the complementary angle that gives the same range out. Level ground, no air resistance, and every trigonometric step taken in radians.

Last updated 11 Nov 2026

Projectile from level ground: range, height, flight time

The horizontal and vertical motions, kept separate

Launch speed u (m/s)The whole speed along the launch direction, not its horizontal part.

Launch angle θ above the horizontal (degrees)Measured up from the ground. 45° gives the longest range on level ground; θ and 90° − θ give the same range.

Value of g to useBoards differ. 9.8, 9.81 and 10 m/s² all appear in school papers, and standard gravity is exactly 9.806 65 m/s² by definition — a conventional value fixed by the 3rd CGPM in 1901, not a measurement. Between 9.8 and 9.81 the answer moves by about one part in a thousand, which is the fourth significant figure of most answers and occasionally rolls the third. Between 9.81 and 10 it moves by 2%, which is visible in the second figure. Use the value your question prints.

59.87mExample

A ball is kicked from level ground at 25 m/s at 35° above the horizontal. Find the range, the maximum height and the time of flight. Take g = 9.81 m/s².

Two independent motions, and the 2θ that makes the range symmetric

T = 2u sin θ / g · H = u²sin²θ / 2g · R = u²sin 2θ / g

u: launch speed, in m/s, along the launch direction.
θ: launch angle above the horizontal. Printed in degrees here and converted to radians before any sine or cosine is taken.
u cos θ: horizontal velocity component, in m/s. Constant for the whole flight, because nothing pushes horizontally.
u sin θ: vertical velocity component at launch, in m/s. This is the only part gravity acts on.
sin 2θ: the factor that makes R(θ) = R(90° − θ), because sin 2θ = sin(180° − 2θ). It peaks at 2θ = 90°, so R peaks at 45°.

Worked example

A ball is kicked from level ground at 25 m/s at 35° above the horizontal. Find the range, the maximum height and the time of flight. Take g = 9.81 m/s².
Resolve first, always. u_x = 25 cos 35° = 25 × 0.86603 = 20.48 m/s. u_y = 25 sin 35° = 25 × 0.57358 = 14.34 m/s. Keep these two apart from here on; mixing them is the error the whole method exists to prevent.
Time of flight from the vertical motion alone. It goes up, stops, and comes back to the same height, so T = 2u_y/g = 2(14.34)/9.81 = 2.923 s. The horizontal motion has nothing to do with this number.
Range from the horizontal motion alone. No horizontal force means constant horizontal velocity, so R = u_xT = 20.48 × 2.9234 = 59.87 m. The one-line form R = u²sin 2θ/g = 625 × sin 70°/9.81 gives the same 59.87 m.
Maximum height from the vertical motion alone. H = u_y²/2g = 205.6/19.62 = 10.48 m, reached at T/2 = 1.462 s.
Two checks worth doing. H/R should equal tan θ/4 = 0.7002/4 = 0.1751, and 10.48/59.87 = 0.1751. And the same 25 m/s at 55° must give the same range, because 35° and 55° are complementary — it does, with a 4.09 s flight and a 21.4 m apex instead.
Radians, not degrees. Every sine and cosine above is of an angle in radians: 35° is 0.610865 rad. Feeding 35 straight into a sine function returns sin(35 radians) = −0.4282, and the range then comes out as 49.30 m — an 18% error, with no warning and no obviously silly figure.

Angle by angle for a 25 m/s launch, with g = 9.81 m/s²

Read the first and last columns together. 15° and 75° give the identical range of 31.86 m, but the 75° shot spends 4.9 s in the air and climbs almost 30 m while the 15° shot is down in 1.3 s having barely cleared 2 m. The range is symmetric about 45°; nothing else is. The H ÷ R column is tan θ/4 in every row, which is a useful one-line check on any projectile answer.
θ	Range R (m)	Max height H (m)	Flight time T (s)	H ÷ R	Same range as
15°	31.86	2.13	1.319	0.0670	75°
30°	55.17	7.96	2.548	0.1443	60°
35°	59.87	10.48	2.923	0.1751	55°
45°	63.71	15.93	3.604	0.2500	itself
60°	55.17	23.89	4.414	0.4330	30°
75°	31.86	29.72	4.923	0.9330	15°
80°	21.79	30.89	5.019	1.4178	10°
90°	0.00	31.86	5.097	—	nothing — the range is zero

Degrees and radians: the conversion that has to be done by hand

Every trigonometric function in physics takes radians, because the radian is the coherent SI unit of plane angle. The last column is what you get if you forget: the values are not absurd, they are not out of range, and several of them are positive, so the wrong answer looks like an answer. For this page’s worked example the range comes out as 49.30 m instead of 59.87 m. Multiply degrees by π/180 first, every time.
Angle in degrees	Angle in radians	sin of it, correctly	sin of the bare number, wrongly
15°	0.261799	0.258819	0.650288
30°	0.523599	0.500000	−0.988032
35°	0.610865	0.573576	−0.428183
45°	0.785398	0.707107	0.850904
60°	1.047198	0.866025	−0.304811
90°	1.570796	1.000000	0.893997

Independence of the two motions, and the symmetry that comes out of sin 2θ

There is only one idea in projectile motion: the horizontal and vertical motions do not talk to each other. Gravity acts straight down, so it changes the vertical velocity and leaves the horizontal velocity completely alone. Resolve the launch velocity into its two components, treat the vertical one as the straight-up-and-down problem and the horizontal one as constant-velocity motion, and then use the time as the only thing the two share.

That is why the method always runs in the same order. Resolve. Find the flight time from the vertical motion, because the vertical motion is what decides when the projectile lands. Multiply that time by the horizontal velocity to get the range. Students who try to find the range directly from a single formula get the right answer on level ground and come apart the moment the ground is not level — which is the launched-from-a-height case, and a different page.

The sin 2θ in the range formula is where the symmetry comes from. Because sin 2θ = sin(180° − 2θ), the angles θ and 90° − θ produce the same range: 20° and 70°, 35° and 55°, 40° and 50°. The maximum is at 2θ = 90°, so 45° gives the longest range on level ground, R = u²/g, and that is also the only angle equal to its own complement. Everything else about the two flights differs — the high one is slower, longer and higher — and an exam question that gives you a range and asks for the angle has two answers unless it also constrains the time or the height.

Where this goes wrong is the radian. Every sine and cosine in physics takes an angle in radians, because the radian is the coherent SI unit of plane angle with dimension one. A launch angle is quoted in degrees, so it has to be multiplied by π/180 before any trigonometric function touches it. The failure mode is quiet: sin(35) interpreted as radians is −0.428 rather than 0.574, the range comes out as 49.30 m rather than 59.87 m, and nothing about 49.30 m looks wrong. The second table above lists the values side by side so you can recognise the symptom. For the record, this calculator does the conversion with an explicit × π/180 rather than a conversion function, because the engine behind these pages has sine and cosine but no degree-to-radian helper at all.

Resolve, then never mix the two again until the end. That is the whole discipline, and the errors it prevents are specific. Using the full launch speed u where the working needs u sin θ inflates the flight time. Applying g to the horizontal motion produces a decelerating horizontal velocity, which nothing causes. Using T rather than T/2 for the climb doubles the maximum height. And looking for the velocity at the apex by setting v = 0 forgets that only the vertical component vanishes. Each of those is a one-line slip that produces a plausible number, so none of them is caught by glancing at the answer — only by keeping the two columns of working physically separate on the page.

The idealisations are worth stating plainly, because they are what separates this page from a real trajectory. No air resistance: a real ball loses range, and the loss is large for a shuttlecock and small for a shot. Level ground: start and finish at the same height, which is what makes the flight symmetric and the apex exactly halfway. Constant g, and no spin, so no swerve and no lift. Each of those is a reason a real sports measurement falls short of the calculated range, and the optimum launch angle for a real thrown object is usually well below 45° once drag is included.

One last check that costs nothing. At any launch angle below 90° the speed at the top of the arc is not zero — only the vertical component is. At the apex the projectile is still travelling horizontally at u cos θ, which for this worked example is 20.48 m/s. “The velocity is zero at the highest point” is true for a vertical throw — the single case θ = 90°, where the horizontal component is zero as well — and false for every projectile launched at any other angle. That difference is a popular exam trap.

Why 45° is worth understanding rather than memorising. The range is the product of the horizontal speed and the flight time, and the launch angle trades one against the other: a flatter launch buys horizontal speed and spends flight time, a steeper one does the reverse. Writing the product out gives the sin 2θ, and a product of two quantities whose sum is fixed is largest when they are equal — which is at 45°, where the two components are both u/√2. Understood that way, the result generalises: anything that breaks the trade, such as launching from a height or paying a drag penalty that grows with time in the air, moves the optimum away from 45° and always downwards.

The path is a parabola, and it is worth knowing why. Eliminate t between the two component equations — x = u cos θ t and y = u sin θ t − ½gt² — and you get y = x tan θ − gx²/(2u²cos²θ), which is a quadratic in x and therefore a parabola. The linear term carries the launch direction and the quadratic term carries the fall. Two consequences are examinable: the apex is at exactly half the range, by the symmetry of a parabola about its vertex, and the path crosses any given height at two values of x, which is the same two-root structure that the vertical motion has in time. A trajectory sketch that is not symmetric about its highest point, on level ground and with no air resistance, is drawn wrong.

Frequently asked questions

Why is 45 degrees the best angle, and is it always?

On level ground, with no air resistance, yes: the range is u²sin 2θ/g and sin 2θ is largest when 2θ = 90°. It stops being true as soon as either idealisation goes. Launch from a height and the best angle falls below 45° — the projectile gets extra fall time for free, so it pays to spend more of the speed going forwards. Add air resistance and the best angle drops further still, which is why real throwers and golfers do not launch at 45°.

Do I use sine or cosine for the vertical component?

Sine, when the angle is measured from the horizontal, which is the usual convention: the vertical component is u sin θ and the horizontal is u cos θ. If a question measures the angle from the vertical instead, the two swap. The safe habit is to sketch the triangle rather than memorise which letter goes where, and to check that the component adjacent to the angle carries the cosine.

Is the velocity zero at the top of the flight?

Not unless you launched at exactly 90°. The vertical component is zero at the top; the horizontal component is u cos θ and has not changed since launch, and it is zero only when θ is 90°. For a 25 m/s launch at 35° the speed at the apex is 20.5 m/s, horizontal. This is one of the most heavily set traps in the topic, and the answer follows straight from the independence of the two motions: nothing horizontal ever changes the horizontal velocity.

Why do two different angles give the same range?

Because sin 2θ = sin(180° − 2θ), so θ and 90° − θ give the same sin 2θ and therefore the same range. 25 m/s at 15° and at 75° both land 31.86 m away. Everything else about the two flights is different: 1.3 s and 2.1 m of height against 4.9 s and 29.7 m. If a question gives a range and asks for an angle, quote both unless something else rules one out.

The calculator gives a range of zero at 90 degrees. Is that right?

Yes. At 90° the horizontal component is zero, so the projectile goes straight up and comes straight back down to the launch point, and the horizontal range is genuinely zero. The flight time and the maximum height stay correct, because they only ever depended on the vertical component. The vertical motion page is the better tool for that case because it also gives the two times at each height.

How much does air resistance actually change this?

Enough to matter, and in a way that depends heavily on the object. The drag force grows with speed, so a fast, light object loses the most: a badminton shuttle falls far short of its calculated range, a football moderately short, a shot put very little. The trajectory also stops being symmetric, with a steeper descent than ascent. No school formula captures this, which is why the question always says to neglect it, and why a measured range in a real experiment comes in under the calculated one rather than scattering either side.

How do I find the velocity partway through the flight, not at the ends?

Treat the two components separately, as always. The horizontal component is u cos θ at every instant, unchanged. The vertical component at time t is u sin θ − gt, which is positive while climbing, zero at the apex and negative afterwards. Combine them with Pythagoras for the speed and with an arctangent for the direction. At t = 2.0 s in the worked example the vertical component is 14.34 − 19.62 = −5.28 m/s, so the ball is already descending at 21.1 m/s, 14.5° below the horizontal.

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References

OCR, Data, Formulae and Relationships Booklet for AS and A Level GCE Physics A (H156 and H556), 8 pages. Lists the acceleration of free fall as g = 9.81 m s⁻² alongside the thirteen other physical constants the papers assume.
BIPM, The International System of Units (SI), 9th edition (2019). Definitions of the metre and the second, and the convention that the radian is the coherent SI unit of plane angle with dimension one — which is why every trigonometric formula in mechanics is written in radians.
NIST, CODATA Internationally Recommended Values of the Fundamental Physical Constants — complete listing. The standard acceleration of gravity is tabulated as 9.806 65 m s⁻², marked exact rather than carrying an uncertainty. physics.nist.gov/cuu/Constants/
NCERT, Physics: Textbook for Class XI, Part I, Chapter 2 “Motion in a Straight Line”, §2.4 “Kinematic Equations for Uniformly Accelerated Motion”. Derives the three equations writing v₀ for the initial velocity and x for the displacement, and takes g as 9.8 m s⁻² (10 m s⁻² in Example 2.3).
BIPM, Declaration on the unit of mass and on the definition of weight; conventional value of g_n — Declaration 2 of the 3rd CGPM (1901), Proceedings p. 70. This is the document that fixes the conventional value at 980.665 cm/s². doi:10.59161/CGPM1901DECL2E