Free Fall vs Projectile Motion: The Difference

Free fall is the special case of projectile motion with no horizontal component. Gravity is the only force in both and the downward acceleration is the same, so a horizontal velocity changes nothing about the fall: a dropped ball and one fired horizontally from the same height land together.

AP Physics: Unit 1 (topics 1.3 Representing Motion, 1.5 Vectors and Motion in Two Dimensions). Both sides of this comparison sit in Unit 1, Kinematics, of AP Physics 1, weighted at 10 to 15 percent of the multiple-choice section. Topic 1.5, Vectors and Motion in Two Dimensions, carries LO 1.5.A on the perpendicular components of a vector (EK 1.5.A.1 through 1.5.A.3) and LO 1.5.B on describing the motion of an object moving in two dimensions, which carries EK 1.5.B.1, that motion in two dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components, and EK 1.5.B.2, that projectile motion is a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension. Suggested skills for Topic 1.5 are 1.B, 2.A, 2.D, 3.A and 3.C, and the topic carries no boundary statement. Topic 1.3, Representing Motion, supplies the vertical column: EK 1.3.A.2 gives the three kinematic equations with a note that they may be used in any single dimension, and EK 1.3.A.3 gives the acceleration caused by gravity as approximately 10 m/s^2, while the Topic 1.3 boundary statement says the exam will use g approximately 10 m/s^2 wherever a numerical value is required and that students will not be penalized for correctly using 9.81 or 9.8 m/s^2. The Table of Information prints g = 9.8 m/s^2, which is the value used throughout this page. The phrase free fall appears nowhere in the AP Physics 1 CED or the AP Physics C: Mechanics CED; both define projectile motion, at EK 1.5.B.2 and EK 1.5.A.4 respectively, in identical words. AP Physics C: Mechanics adds EK 1.5.A.3, that motion in one dimension may be changed without causing a change in a perpendicular dimension, which AP Physics 1 uses but never states. No AP equation sheet prints a range or time-of-flight equation.

One motion, seen with and without a sideways push

Most pages treat these as two topics with two method boxes. They are one motion.

AP Physics 1 EK 1.5.B.2 defines projectile motion as a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension. Set the horizontal velocity to zero and that description still holds word for word. The horizontal dimension still has zero acceleration; it just also has zero velocity, so nothing happens in it. What is left is a straight vertical drop, which is what people call free fall.

So the honest statement of the difference is a statement about a number, not about physics:

Free fall: $v_{x0} = 0$ .
Projectile motion: $v_{x0}$ is whatever the launch gave it, and may be zero.

Everything in the vertical column, the acceleration, the equations, the time to fall, the speed gained per second, is the same on both sides, and that is the exam-relevant consequence. The horizontal component is along for the ride.

The word "projectile" is not a shape or a size. It is a condition: gravity alone acts. A dropped stone qualifies, a kicked ball qualifies, a rocket with its engine burning does not. The projectile and free fall glossary entries define each side in a sentence; this page is about what happens when you put them next to each other.

Free fall vs projectile motion, side by side

Question you are asking	Free fall	Projectile motion
Forces acting	Gravity only	Gravity only
Horizontal acceleration	$0$ , and $v_x = 0$ too	$0$ , with $v_x$ constant and nonzero
Vertical acceleration	$9.8\ \text{m/s}^2$ downward	$9.8\ \text{m/s}^2$ downward
Path shape	A straight vertical line	A parabola
Number of axes you must set up	One	Two
Time to fall a given height	$t = \sqrt{2h/g}$	$t = \sqrt{2h/g}$ , the same number
Does the launch speed change the fall time	There is no launch speed	No, only the vertical part does
Speed at landing	The vertical speed alone	$\sqrt{v_x^2 + v_y^2}$ , larger
Equations used	The three kinematic equations, once	The same three, once per axis
Where the CED puts it	No separate topic	Topic 1.5, EK 1.5.B.2
Term used in the CED	Not used at all	"Projectile motion", defined

Three rows carry the whole page.

The two acceleration rows are identical, and the path row is not. A parabola and a vertical line look nothing alike, which is why the two get taught as separate lessons. But the shape is a consequence of the horizontal component, not of any difference in what gravity is doing.

The fall-time row is the exam consequence. $t = \sqrt{2h/g}$ contains no $v_x$ . There is nowhere for a horizontal velocity to enter, so it cannot change the answer. Worked example one puts a number on that.

The last two rows are a vocabulary warning worth having. The phrase "free fall" appears nowhere in the AP Physics 1 Course and Exam Description, and nowhere in the AP Physics C: Mechanics one either. Both define projectile motion, in identical words, at EK 1.5.B.2 and EK 1.5.A.4 respectively. The idea of free fall is everywhere in those courses; the label is not, so an exam question will describe the situation rather than name it.

The case that separates them: drop one, fire the other

Hold two identical balls at the same height. Drop one. At the same instant, fire the other horizontally at $6.0$ m/s. They hit the floor at the same moment.

This is the experiment the whole comparison rests on, and it is worth being exact about why it works. The reason is not that the horizontal motion is small or slow. It is that the two axes are solved by separate equations that share only the clock.

AP Physics 1 EK 1.5.B.1 grants the split: motion in two dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components. The AP Physics C: Mechanics CED states the underlying fact outright at EK 1.5.A.3: motion in one dimension may be changed without causing a change in a perpendicular dimension. That sentence has no counterpart in the AP Physics 1 CED, which grants the method without stating the principle, so a Physics 1 student is expected to use the independence without ever seeing it written down.

Worked example one runs both balls from $1.5$ m. Both are in the air for $0.55$ s. The fired ball travels $3.3$ m sideways in that time and lands with a larger speed, because it kept the horizontal velocity it started with while gaining exactly the same vertical velocity as the dropped ball.

Double the launch speed to $12$ m/s and the fired ball lands twice as far away, at the same instant, with the same downward velocity component. The horizontal column can be changed freely and the vertical column does not notice.

What free fall does not mean

Two readings of the phrase cause more lost marks than the comparison itself does.

Free fall does not mean falling. It describes which forces act, not which way the object is going. A ball thrown straight upward is in free fall from the instant it leaves the hand: on the way up, at the top, and on the way down. Gravity is the only force throughout, so the acceleration is $9.8\ \text{m/s}^2$ downward at every one of those moments. At the top the velocity is zero and the acceleration is not.

That also means a projectile launched at an angle is in free fall for its whole flight, upward leg included. There is no moment when it stops being one thing and starts being another.

Free fall does not mean weightless, and it does not mean no air. It means the model has been set up so that gravity is the only force in it. The AP Physics 1 exam conventions grant that setup as a standing assumption: air resistance is assumed to be negligible unless otherwise stated. That single line is what keeps the horizontal velocity constant, and it is why a real ball fired horizontally does drift behind the ideal answer.

When air resistance is not negligible, the object is no longer in free fall and neither the parabola nor the equal-fall-time result survives. That regime belongs to AP Physics C: Mechanics Topic 2.9, where a resistive force makes the vertical motion exponential and produces a terminal velocity. Nothing like it appears in AP Physics 1.

Where the confusion costs a mark

Each of these is a specific scoring error, and all of them come from treating the two as separate physics.

Using the launch speed where the vertical component belongs. For a launch at $30^\circ$ with speed $18$ m/s, the time in the air is set by $v_{y0} = 9.0$ m/s, not by $18$ m/s. Substituting the full speed doubles the flight time.
Thinking the faster projectile stays up longer. Worked example three fires the same $18$ m/s at $30^\circ$ and at $60^\circ$ . The ranges match to the digits printed. The flight times differ by a factor of $\sqrt{3}$ , because only the vertical component sets the clock.
Setting the acceleration to zero at the top. The velocity is zero there; the acceleration is $9.8\ \text{m/s}^2$ downward. An object with zero acceleration at the top would hang there.
Applying $9.8$ to the horizontal axis. $a_x = 0$ for the entire flight. Any horizontal equation with a $g$ in it is wrong before the arithmetic starts.
Solving the horizontal axis first. Only the vertical column contains enough information to find $t$ . Find $t$ there, then hand it across.
Treating a dropped object as a different problem. It is a projectile with $v_{x0} = 0$ , and the same three equations solve it with one column left blank.
Forgetting that landing speed is not landing vertical speed. A fired ball lands faster than a dropped one from the same height, because $v_x$ never went away. In worked example one the difference is $5.4$ m/s against $8.1$ m/s.
Assuming free fall has a terminal velocity. In free fall the acceleration never changes, so the speed grows without limit in the model. A terminal speed requires a second force, which by definition takes the object out of free fall.

The step-by-step routine for both belongs to how to solve projectile motion problems, and the projectile motion calculator will check a number against it.

Which g the exam actually uses

This comparison is where the $g$ question surfaces, because both sides depend on the same value and a page that quietly used two of them would produce two different answers to one question.

AP Physics 1 EK 1.3.A.3 states that near the surface of Earth, the vertical acceleration caused by the force of gravity is downward, constant, and has a measured value approximately equal to $a_g = g \approx 10\ \text{m/s}^2$ . The Topic 1.3 boundary statement then commits the exam: for all situations in which a numerical quantity is required for $g$ , the value $g \approx 10\ \text{m/s}^2$ will be used. The same boundary statement continues, and the continuation is the part that gets dropped when it is quoted: students will not be penalized for correctly using the more precise commonly accepted values of $g = 9.81\ \text{m/s}^2$ or $g = 9.8\ \text{m/s}^2$ .

The AP Physics 1 Table of Information prints $g = 9.8\ \text{m/s}^2$ in its constants box, alongside $g = 9.8\ \text{N/kg}$ for the field strength. So both values are real and both are sanctioned. This site computes with $9.8$ throughout, including every number on this page.

What matters for the comparison is that the choice cancels out of the thing being compared. The dropped ball and the fired ball share whatever value you pick, so they land together at $9.8$ and they land together at $10$ . Only the individual times change. The full treatment of which to use and when is in is g 9.8 or 10 in AP Physics 1.

What the CED asks, and how the exam frames it

Both sides of this comparison live in Unit 1, Kinematics, weighted at 10 to 15 percent of the AP Physics 1 multiple-choice section.

Topic 1.5, Vectors and Motion in Two Dimensions, carries two learning objectives. LO 1.5.A asks you to describe the perpendicular components of a vector, with EK 1.5.A.1 through 1.5.A.3 supplying the resolution into components and the trigonometric relationships. LO 1.5.B asks you to describe the motion of an object moving in two dimensions, and carries exactly two pieces of essential knowledge: EK 1.5.B.1, that motion in two dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components, and EK 1.5.B.2, the definition of projectile motion. Suggested skills for Topic 1.5 are 1.B, 2.A, 2.D, 3.A and 3.C. The topic carries no boundary statement.

Topic 1.3, Representing Motion, is where the vertical column comes from. EK 1.3.A.2 gives the three kinematic equations for constant acceleration and adds a note that they are written for the x-direction but can be used in any single dimension as appropriate, which is the licence to run them down the y-axis. EK 1.3.A.3 gives the value of the gravitational acceleration, and the topic's two boundary statements cover nonuniform acceleration and the value of $g$ .

On the equation sheet, the three kinematic equations appear once, in the $x$ form, and that is all you get. There is no range equation and no time-of-flight equation printed on any of the four AP sheets. Both are derived on the page from the same three lines, which is another way of saying the exam expects you to treat a projectile as two one-dimensional problems rather than to recall a projectile formula.

AP Physics C: Mechanics covers the same ground in its own Topic 1.5, Motion in Two or Three Dimensions, with EK 1.5.A.4 giving the identical definition of projectile motion and EK 1.5.A.3 adding the independence statement. Its boundary statement limits quantitative analysis to two dimensions and hands three-dimensional description to AP Physics C: Electricity and Magnetism.

For practice, the projectile motion set and the kinematics set both work these; the CED framing is on Topic 1.5 and Topic 1.3.

Dropped and fired, from the same height

Two identical balls start at rest at a height of $1.5$ m above the floor. Ball A is released from rest. At the same instant Ball B is launched horizontally at $6.0$ m/s. Find the time each is in the air, how far Ball B lands from the launch point, and the landing speed of each. Then repeat the horizontal distance for a launch at $12$ m/s. Take downward as positive for the vertical axis, use $g = 9.8\ \text{m/s}^2$ and ignore air resistance.

Set the axes and declare the sign convention before anything else: $y$ positive downward, $x$ positive in the direction of launch. With that choice $a_y = +9.8\ \text{m/s}^2$ and $a_x = 0$ for both balls.
Ball A, the vertical column. Starting from rest, $y = v_{y0}t + \frac{1}{2}a_y t^2$ becomes $1.5 = \frac{1}{2}(9.8)t^2$ , so $t^2 = \frac{2(1.5)}{9.8} = 0.30612$ and $t = 0.5533\ \text{s}$ .
Ball B, the vertical column. Ball B was launched horizontally, so its initial vertical velocity is also zero. The equation is character for character the same one, and returns the same $t = 0.5533\ \text{s}$ . Nothing about the launch entered it: $v_{x0}$ does not appear in any vertical equation.
Ball B, the horizontal column. $a_x = 0$ , so $x = v_{x0}t = (6.0)(0.5533) = 3.32\ \text{m}$ . That is the only place the launch speed is used in the entire problem.
Landing speeds. Both balls gain the same vertical velocity: $v_y = a_y t = (9.8)(0.5533) = 5.42\ \text{m/s}$ downward. Ball A has nothing else, so it lands at $5.42\ \text{m/s}$ .
Ball B still has the $6.0\ \text{m/s}$ it started with, because nothing acted horizontally. Its landing speed is $\sqrt{(6.0)^2 + (5.42)^2} = \sqrt{36.0 + 29.4} = \sqrt{65.4} = 8.09\ \text{m/s}$ , at $42^\circ$ below the horizontal.
Now double the launch speed. $t$ is unchanged at $0.5533\ \text{s}$ , because the vertical equation was never touched. The range becomes $x = (12)(0.5533) = 6.64\ \text{m}$ , exactly twice as far, landing at exactly the same instant.
Line them up. Fall time: $0.5533$ s, $0.5533$ s, $0.5533$ s. Vertical landing velocity: $5.42$ m/s in all three cases. Horizontal distance: $0$ , $3.32$ m, $6.64$ m. The vertical column is a constant across the table and the horizontal column is free.

Both balls fall for $0.55$ s. Ball B lands $3.32$ m away at $8.09$ m/s; Ball A lands beneath its start at $5.42$ m/s. At $12$ m/s the range doubles to $6.64$ m with the fall time unchanged. The only quantity the launch speed changed was the horizontal distance.

Straight up against a launch at an angle, with the same vertical component

Ball P is thrown straight upward at $12$ m/s from ground level. Ball Q is launched from the same spot at $20$ m/s at $37^\circ$ above the horizontal. Using the CED trigonometry table values $\sin 37^\circ = 0.6$ and $\cos 37^\circ = 0.8$ , find the initial vertical component of each, the time each spends in the air, the maximum height each reaches, and the horizontal range of each. Take upward as positive and use $g = 9.8\ \text{m/s}^2$ .

Resolve Ball Q. $v_{y0} = (20)(0.6) = 12.0\ \text{m/s}$ and $v_{x0} = (20)(0.8) = 16.0\ \text{m/s}$ . Ball P has $v_{y0} = 12.0\ \text{m/s}$ and $v_{x0} = 0$ . The two balls have identical vertical columns and different horizontal ones, which is the whole setup.
Time to the top, from $v_y = v_{y0} - gt$ with $v_y = 0$ : $t = \frac{12.0}{9.8} = 1.224\ \text{s}$ . Both balls, same equation, same number.
Total time in the air. The flight is symmetric about the peak for a launch and landing at the same height, so $t_{\text{total}} = 2(1.224) = 2.449\ \text{s}$ for both.
Maximum height, from $v_y^2 = v_{y0}^2 - 2g\Delta y$ with $v_y = 0$ : $\Delta y = \frac{(12.0)^2}{2(9.8)} = \frac{144}{19.6} = 7.35\ \text{m}$ . Both balls, again the same number. Ball Q is $19.6$ m downrange when it gets there, and that has no bearing on the height.
Ranges. Ball P: $x = (0)(2.449) = 0$ , it lands on the thrower's head. Ball Q: $x = (16.0)(2.449) = 39.2\ \text{m}$ .
Check the state at the top for both. Ball P is momentarily at rest, velocity zero. Ball Q is moving at $16.0\ \text{m/s}$ horizontally, velocity not zero. Both have an acceleration of $9.8\ \text{m/s}^2$ downward at that instant, because gravity is the only force on either. A question asking for the acceleration at the peak has the same answer for the two.
The difference in the two flights is one number, $v_{x0}$ , and it changed exactly one answer, the range.

Both have $v_{y0} = 12.0$ m/s, both are in the air $2.45$ s, and both reach $7.35$ m. Ball P has a range of zero and Ball Q a range of $39.2$ m. At the peak both have an acceleration of $9.8\ \text{m/s}^2$ downward, though Ball P is momentarily at rest and Ball Q is doing $16.0$ m/s.

Equal ranges, unequal times

A ball is launched from level ground at $18$ m/s, once at $30^\circ$ above the horizontal and once at $60^\circ$ . Find the flight time, range and maximum height for each. Take upward as positive and use $g = 9.8\ \text{m/s}^2$ .

Components at $30^\circ$ : $v_{x0} = 18\cos 30^\circ = 15.59\ \text{m/s}$ , $v_{y0} = 18\sin 30^\circ = 9.00\ \text{m/s}$ , using $\sin 30^\circ = 1/2$ and $\cos 30^\circ = \sqrt{3}/2$ from the trigonometry table printed on the sheet.
Components at $60^\circ$ : $v_{x0} = 18\cos 60^\circ = 9.00\ \text{m/s}$ , $v_{y0} = 18\sin 60^\circ = 15.59\ \text{m/s}$ . The two component pairs are the same two numbers swapped.
Flight times, from $t = \frac{2v_{y0}}{g}$ . At $30^\circ$ : $t = \frac{2(9.00)}{9.8} = 1.837\ \text{s}$ . At $60^\circ$ : $t = \frac{2(15.59)}{9.8} = 3.181\ \text{s}$ . The steeper launch stays up $\sqrt{3}$ times as long, and the ratio is exactly the ratio of the vertical components.
Ranges, $x = v_{x0}t$ . At $30^\circ$ : $(15.59)(1.837) = 28.6\ \text{m}$ . At $60^\circ$ : $(9.00)(3.181) = 28.6\ \text{m}$ . Equal. Each launch trades horizontal speed against time in the air in exactly compensating proportion.
Maximum heights, $\Delta y = \frac{v_{y0}^2}{2g}$ . At $30^\circ$ : $\frac{81.0}{19.6} = 4.13\ \text{m}$ . At $60^\circ$ : $\frac{243}{19.6} = 12.4\ \text{m}$ , three times as high.
Read what this proves. Two flights with the same range, the same launch speed and the same landing point differ by a factor of $1.73$ in time and $3$ in height. Range on its own tells you nothing about the vertical story, and time in the air tells you nothing about the horizontal one. They are separate columns that happen to share a clock.

At $30^\circ$ : $t = 1.84$ s, range $28.6$ m, peak $4.13$ m. At $60^\circ$ : $t = 3.18$ s, range $28.6$ m, peak $12.4$ m. Same launch speed and same range, with the flight time and peak height set entirely by the vertical component.

Frequently asked questions

Is free fall the same as projectile motion?

Free fall is the special case of projectile motion in which the horizontal velocity is zero. They are not two different phenomena. In both, gravity is the only force acting, the horizontal acceleration is zero and the vertical acceleration is 9.8 metres per second squared downward near Earth's surface. The AP Physics 1 CED defines projectile motion at essential knowledge 1.5.B.2 as a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension, and that description still holds when the horizontal velocity happens to be zero. The only difference is the shape of the path: a straight vertical line rather than a parabola.

Do a dropped ball and a ball fired horizontally hit the ground at the same time?

Yes, if they start at the same height and air resistance is negligible. The time to fall comes from the vertical equations alone, and the horizontal velocity does not appear in any of them. From 1.5 metres both take 0.55 seconds, whether the second ball is fired at 6 metres per second, 12 metres per second or not at all. Doubling the launch speed doubles the horizontal distance and leaves the fall time untouched. The AP Physics C: Mechanics CED states the principle directly at essential knowledge 1.5.A.3: motion in one dimension may be changed without causing a change in a perpendicular dimension.

Is an object at the top of its flight still in free fall?

Yes. Free fall describes which forces act, not which way the object is moving. A ball thrown straight upward is in free fall from the instant it leaves the hand, through the peak, and all the way down, because gravity is the only force on it throughout. At the very top the velocity is momentarily zero and the acceleration is still 9.8 metres per second squared downward. Those two facts are separate: zero velocity does not imply zero acceleration. If the acceleration really were zero at the peak, the ball would stay there.

Does the AP Physics 1 CED use the term free fall?

No. The phrase does not appear anywhere in the AP Physics 1 Course and Exam Description, or in the AP Physics C: Mechanics one. Both define projectile motion instead, in identical wording at essential knowledge 1.5.B.2 and 1.5.A.4. The physics of free fall is required content, through essential knowledge 1.3.A.3 on the acceleration caused by gravity and the kinematic equations of Topic 1.3, but an exam question will describe the situation rather than use the label. Free fall remains standard physics vocabulary and is worth knowing; just do not expect the exam to prompt you with it.

Does the mass of a projectile affect its motion?

No, not while gravity is the only force acting. The gravitational force grows in proportion to the mass and the resistance to being accelerated grows with exactly the same mass, so the two cancel and every object gets the same 9.8 metres per second squared downward. A marble and a bowling ball released together fall together, and two projectiles launched identically follow the same parabola whatever they weigh. Mass would matter if air resistance were included, because the drag force does not scale with mass, but the AP Physics 1 exam conventions state that air resistance is assumed to be negligible unless a question says otherwise.

Why does a projectile launched at 30 degrees have the same range as one at 60 degrees?

Because the two launches swap their velocity components. At the same launch speed, the 30 degree launch has the larger horizontal component and the smaller vertical one, and the 60 degree launch has them the other way round. Range is horizontal speed multiplied by time in the air, and time in the air is set by the vertical component, so a bigger horizontal speed for a shorter time gives the same product as a smaller horizontal speed for a longer time. At 18 metres per second both land 28.6 metres away, but the 60 degree launch is in the air 3.18 seconds against 1.84 seconds and peaks at 12.4 metres against 4.13 metres.

Does a projectile have a terminal velocity?

Not in the AP Physics 1 model. Terminal velocity requires a resistive force that grows with speed until it balances the weight, and that only exists once air resistance is included, at which point the object is no longer in free fall and its path is no longer a parabola. AP Physics 1 assumes air resistance is negligible unless a question says otherwise, so the downward speed grows without limit in its model. Terminal velocity is developed only in AP Physics C: Mechanics, at Topic 2.9, where essential knowledge 2.9.A.3 defines it as the maximum speed reached when a constant force and a resistive force in opposite directions leave zero net force.