Free Fall and Vertical Motion Calculator

Q: Why do I get two answers for the time?

Because the equation is a quadratic in t and the object really is at that height twice. Thrown upwards at 20 m/s it passes 10 m at 0.584 s going up and again at 3.494 s coming down. The speed is identical both times — 14.28 m/s — and only the direction differs. Which root the question wants depends on its wording, and a question that just asks when the object is at that height is asking for both.

Q: Is the acceleration zero at the highest point?

No. The velocity is zero at the highest point; the acceleration is −g there exactly as it is everywhere else on the flight. If the acceleration were zero the object would hover. This confusion is worth naming because it is the most common wrong statement in the whole topic: velocity and acceleration are independent quantities and either can be zero while the other is not.

Q: What value of g should I use?

The one printed on your own paper. Standard gravity is exactly 9.806 65 m/s² by definition, fixed by the 3rd CGPM in 1901 and listed as exact in the NIST constants table, but that is a conventional value rather than the local one. NCERT worked examples use 9.8; the OCR, CIE and Edexcel data booklets print 9.81; some papers permit 10 for mental arithmetic. The table above shows all four applied to the same throw — and shows that the 9.8-versus-9.81 difference can move a three-figure answer, so it is not purely cosmetic.

Q: How do I handle an object dropped from a height rather than thrown?

Set u = 0 and make h negative, because the landing point is below the release point. The quadratic loses its linear term and reduces to t = √(2|h|/g). A fall of 45 m takes 3.029 s with g = 9.81. The mass of the object does not appear, which is the point of the experiment: with air resistance neglected, everything falls at the same rate.

Q: My object is thrown downwards. Do I make u negative or h negative?

Both, usually. With up positive, a downward throw has a negative u, and a target below the launch point has a negative h. Typing a positive u for a downward throw is the commonest sign error on this page and it produces a plausible-looking answer, which is what makes it dangerous. Check it by eye: a downward throw must reach a point below it sooner than a plain drop would.

Q: Why does the calculator sometimes refuse to give an answer?

Because the height you asked about is above the apex of the throw, so the object never reaches it. The discriminant u² − 2gh goes negative, the roots become complex, and there is no time at which the object is at that height. The highest point a launch speed u can reach is u²/2g — 20.4 m for 20 m/s with g = 9.81 — and asking about 25 m gives no answer rather than a wrong one.

Q: How do I check my two roots are right without redoing the algebra?

Compare them with the apex time, u/g. A root on the way up must be smaller than it and a root on the way down must be larger, so the two roots have to straddle it — and their mean must equal it exactly, because the quadratic is symmetric about its vertex. For the worked example 0.584 and 3.494 have a mean of 2.039, which is u/g. If both your roots land on the same side of the apex time, you have a sign error in the discriminant.

Throw something straight up and it passes every height twice. s = ut − ½gt² is a quadratic in t, so it has two roots, and deciding which one the question wants is the mark. This page solves it, prints both roots, and tells you when there is no root at all.

Last updated 11 Nov 2026

Vertical motion under gravity: the quadratic and both its roots

Two roots, the apex, and the case with no answer

Launch velocity u, upwards positive (m/s)Positive for a throw upwards, zero for a release from rest, negative for a throw downwards. Up-positive is a choice, not a law; it is the one this page makes, and it is why the acceleration comes out as −g.

Target height h above the launch point (m)Zero is the launch point itself. Use a negative number for a point below it — the foot of a cliff, the ground under a balcony, the bottom of a well.

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.

0.584sExample

A ball is thrown vertically upwards at 20 m/s. When is it 10 m above the throwing hand? Take g = 9.81 m/s².

One quadratic, two roots, and the discriminant that is a speed in disguise

h = ut − ½gt² → ½gt² − ut + h = 0 → t = (u ± √(u² − 2gh)) / g

u: launch velocity, in m/s, positive upwards. Zero for a release from rest.
h: height of the target point above the launch point, in m. Negative below it.
g: magnitude of the free-fall acceleration, in m/s². The acceleration itself is −g, because up was chosen as positive.
u² − 2gh: the discriminant — and also the square of the speed at height h, by v² = u² − 2gh. Negative means the height is never reached; zero means it is reached exactly at the apex.
t: time after launch, in s. Two roots; a negative one is unphysical for a motion that starts at t = 0.

Worked example

A ball is thrown vertically upwards at 20 m/s. When is it 10 m above the throwing hand? Take g = 9.81 m/s².
Set up the quadratic, keeping up as positive. h = ut − ½gt² becomes 10 = 20t − 4.905t², so 4.905t² − 20t + 10 = 0.
Compute the discriminant first. u² − 2gh = 400 − 2(9.81)(10) = 400 − 196.2 = 203.8 m²/s². It is positive, so the ball does reach 10 m — and √203.8 = 14.28 m/s is the speed it has as it passes that height.
Both roots. t = (20 ± 14.276)/9.81, giving t = 0.584 s and t = 3.494 s. The first is on the way up, the second on the way down, 2.91 s later.
Sanity-check against the apex. The top is at u²/2g = 400/19.62 = 20.39 m, reached at u/g = 2.039 s. Our two roots straddle 2.039 s, as they must, and 10 m is comfortably below 20.39 m.
Read the question again. “When is it 10 m above the hand” has two answers and a full-marks response gives both. “When does it first reach 10 m” has one. Candidates lose marks here far more often than they lose them on the arithmetic.
And check what g did. With g = 9.8 the first root is 0.5834 s; with 9.81 it is 0.5835 s; with the exact standard 9.806 65 it is 0.5835 s. To three significant figures that is 0.583, 0.584, 0.583 — so the choice of g moves this answer across a rounding boundary at the third figure. With g = 10 it is 0.586 s.

What the choice of g changes, for this throw to this height

Between 9.8 and 9.81 the answers differ by about one part in a thousand — and that is enough to move *this* first root from 0.583 s to 0.584 s once it is rounded to three significant figures, because it happens to sit near a rounding boundary. So the usual advice that the difference only shows in the fourth figure is right on average and wrong often enough to matter. Use the value printed on your own paper. g = 10 is a different kind of choice: it changes the answer by about 2%, in the second figure, and is only ever intended for mental arithmetic.
g (m/s²)	Where it comes from	First root (s)	Second root (s)	Highest point (m)
9.8	NCERT worked examples, most CBSE marking	0.583	3.498	20.408
9.81	OCR, CIE and Edexcel data booklets	0.584	3.494	20.387
9.80665	standard gravity, exact by definition (CGPM 1901)	0.583	3.495	20.394
10	the quick estimate some papers allow	0.586	3.414	20.000

The four vertical cases, and how the two roots behave in each

The discriminant does all the deciding, and it is not an abstract algebraic quantity here: u² − 2gh is the square of the speed the object has at height h. A negative discriminant means an imaginary speed, which is the algebra’s way of saying the object is never at that height at all.
Case	u	h	Discriminant	Roots	What to take
Thrown up, target below the apex	positive	positive	positive and less than u²	two positive	both are real; the question decides
Thrown up, target exactly at the apex	positive	= u²/2g	zero	one repeated root, at u/g	the single time u/g
Thrown up, target above the apex	positive	> u²/2g	negative	none real	it never gets there
Released from rest, target below	zero	negative	−2gh, positive	one positive, one negative	the positive one, √(2\|h\|/g)
Thrown downwards, target below	negative	negative	larger than u²	one positive, one negative	the positive one
Thrown up, target at the launch point	positive	zero	exactly u²	t = 0 and t = 2u/g	the second one, which is the time of flight
Released from rest, target above	zero	positive	−2gh, negative	none real	it never rises at all

Why a height has two times, and the signs that decide the rest

Vertical motion is not a separate topic. It is the equations of motion with one substitution: a = −g, because gravity pulls down and this page has chosen up as positive. Every result below follows from that one line. What makes it worth its own page is the quadratic: as soon as you are given a height and asked for a time, you are solving ½gt² − ut + h = 0, and that has two roots.

Both roots are usually real and usually positive, and that is not a defect of the maths. A ball thrown upwards genuinely is at 10 m twice: once going up, once coming down. The arithmetic knows this and the question often does not say which one it wants. Read the verb: first reaches is the smaller root, passes on the way down is the larger, and a bare when is it at wants both. Marks are lost here far more often than on the algebra.

The discriminant is a speed. This is the most useful thing on the page and almost no textbook says it out loud. The quantity under the square root, u² − 2gh, is exactly v² at height h, from the velocity–displacement equation. So the square root is the speed the object has as it passes that height, and the ± in the quadratic formula is the ± on the direction of that velocity: upwards the first time, downwards the second, same magnitude both times. If the discriminant is negative, the required speed would be imaginary, which is the algebra saying the object never gets there.

Signs, and the one that catches everybody. With up positive: a throw upwards has u > 0, the ground below the launch point has h < 0, and the acceleration is −g whether the object is rising, stationary at the top, or falling. The acceleration does not become zero at the highest point. The velocity does. Those are different quantities, and “at the top the acceleration is zero” is probably the single most common wrong sentence written about this topic. If it were zero the object would stay there.

Which g to use is a real question, not a pedantic one. Standard gravity is fixed by definition at exactly 9.806 65 m/s² — a conventional value adopted by the 3rd CGPM in 1901 and still carried in the NIST constants table as exact rather than measured. It is not the local value anywhere: true g varies with latitude and altitude by a few parts per thousand. School papers round it to 9.8, 9.81 or 10, and the table above shows what each choice does to one specific answer. The rule is simple: use the number your own question prints, and say which you used.

The symmetry of a throw from and back to the same height is worth stating as a set of results, because questions lean on it constantly. The time up equals the time down, so the total flight time is 2u/g and the apex is at exactly u/g. The speed at any height on the way down equals the speed at that height on the way up — which falls straight out of the discriminant being v² at that height, with the ± only changing the direction. And the landing speed back at the launch point equals the launch speed. All four of those break the moment the start and finish heights differ, which is why a throw from a cliff is a genuinely different problem and gets its own page.

Two sanity checks that cost a line each. First, compare your time against the apex time u/g: a root on the way up must be smaller than it and a root on the way down must be larger, and if both your roots land on the same side of it you have made a sign error. Second, compare the height you asked about against the apex height u²/2g: if it is larger there should be no answer at all, and if an answer appears anyway you have dropped a minus sign somewhere in the discriminant. Both checks take seconds and both catch the errors that otherwise survive to the end of a question.

Air resistance is ignored throughout, as every school treatment ignores it. That is a real approximation, not a formality: a dropped tennis ball reaches a terminal speed and a dropped sheet of paper never follows these equations at all. What saves the model is that the resistance force depends on speed, so for the first second or two of a fall from rest it is small compared with the weight. A question that mentions a parachute, a feather or a long drop is telling you the model has stopped applying.

What the mass does not do is worth as much as what g does. Nothing in any equation on this page contains a mass. A cannonball and a marble released together reach the ground together, because the force of gravity is proportional to mass and so is the inertia that resists it, and the two cancel exactly. That cancellation is not a coincidence of arithmetic — it is the equivalence of gravitational and inertial mass, which is measured to extraordinary precision and is the starting point of general relativity. For school purposes the practical consequence is simply that a question giving you the mass of a falling object is either giving you information you do not need, or is about to ask for a force or an energy rather than a time.

Frequently asked questions

Why do I get two answers for the time?

Because the equation is a quadratic in t and the object really is at that height twice. Thrown upwards at 20 m/s it passes 10 m at 0.584 s going up and again at 3.494 s coming down. The speed is identical both times — 14.28 m/s — and only the direction differs. Which root the question wants depends on its wording, and a question that just asks when the object is at that height is asking for both.

Is the acceleration zero at the highest point?

No. The velocity is zero at the highest point; the acceleration is −g there exactly as it is everywhere else on the flight. If the acceleration were zero the object would hover. This confusion is worth naming because it is the most common wrong statement in the whole topic: velocity and acceleration are independent quantities and either can be zero while the other is not.

What value of g should I use?

The one printed on your own paper. Standard gravity is exactly 9.806 65 m/s² by definition, fixed by the 3rd CGPM in 1901 and listed as exact in the NIST constants table, but that is a conventional value rather than the local one. NCERT worked examples use 9.8; the OCR, CIE and Edexcel data booklets print 9.81; some papers permit 10 for mental arithmetic. The table above shows all four applied to the same throw — and shows that the 9.8-versus-9.81 difference can move a three-figure answer, so it is not purely cosmetic.

How do I handle an object dropped from a height rather than thrown?

Set u = 0 and make h negative, because the landing point is below the release point. The quadratic loses its linear term and reduces to t = √(2|h|/g). A fall of 45 m takes 3.029 s with g = 9.81. The mass of the object does not appear, which is the point of the experiment: with air resistance neglected, everything falls at the same rate.

My object is thrown downwards. Do I make u negative or h negative?

Both, usually. With up positive, a downward throw has a negative u, and a target below the launch point has a negative h. Typing a positive u for a downward throw is the commonest sign error on this page and it produces a plausible-looking answer, which is what makes it dangerous. Check it by eye: a downward throw must reach a point below it sooner than a plain drop would.

Why does the calculator sometimes refuse to give an answer?

Because the height you asked about is above the apex of the throw, so the object never reaches it. The discriminant u² − 2gh goes negative, the roots become complex, and there is no time at which the object is at that height. The highest point a launch speed u can reach is u²/2g — 20.4 m for 20 m/s with g = 9.81 — and asking about 25 m gives no answer rather than a wrong one.

How do I check my two roots are right without redoing the algebra?

Compare them with the apex time, u/g. A root on the way up must be smaller than it and a root on the way down must be larger, so the two roots have to straddle it — and their mean must equal it exactly, because the quadratic is symmetric about its vertex. For the worked example 0.584 and 3.494 have a mean of 2.039, which is u/g. If both your roots land on the same side of the apex time, you have a sign error in the discriminant.

Related calculators

References

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/
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
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).
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.