2.5 Free-Fall Acceleration: What Is g = 9.8 m/s²?

If you throw anything straight up — a tomato, a baseball, a feather in a vacuum chamber — its motion after it leaves your hand is not powered by you anymore. It is governed by a single, constant, downward acceleration that is the same for every object, and it is the most tested constant-acceleration case in every serious physics exam. Mastering free-fall acceleration means you can solve any up-and-down flight time, maximum height, and two-pass height problem for AP Physics, A-Level, IB HL, and Olympiad mechanics without hesitation. In this guide we rebuild the entire topic from the source of truth in Fundamentals of Physics, with zero omission, so you understand why we switch from x to y, why acceleration is always -g, and why a light feather and a heavy apple fall together.

Key Insights

• Free-fall is motion with constant downward acceleration g when air resistance is negligible.
• The magnitude of free-fall acceleration is \(g = 9.8\,\text{m/s}^2\) and is independent of mass.
• In free-fall we use a vertical \(y\)-axis with \(+y\) up and replace \(a\) with \(-g\).
• Velocity decreases on the way up, is zero at the top, and increases downward in the negative direction.
• The same height is passed twice, giving two valid times from the quadratic equation.

Free-Fall Acceleration

Learning Objectives

After this module, you should be able to identify that if a particle is in free flight, whether upward or downward, and if we can neglect the effects of air on its motion, the particle has a constant downward acceleration with a magnitude \(g\) that we take to be \(9.8\,\text{m/s}^2\). You should also be able to apply the constant-acceleration kinematic equations to free-fall motion.

Key Idea

An important example of straight-line motion with constant acceleration is that of an object rising or falling freely near Earth's surface. The constant-acceleration equations describe this motion, but we make two changes in notation: (1) We refer the motion to the vertical \(y\) axis with \(+y\) vertically up; (2) we replace \(a\) with \(-g\), where \(g\) is the magnitude of the free-fall acceleration. Near Earth's surface, \(g = 9.8\,\text{m/s}^2 = 32\,\text{ft/s}^2\).

Free-Fall Acceleration

If you tossed an object either up or down and could somehow eliminate the effects of air on its flight, you would find that the object accelerates downward at a certain constant rate. That rate is called the free-fall acceleration, and its magnitude is represented by \(g\). The acceleration is independent of the object's characteristics, such as mass, density, or shape; it is the same for all objects.

This is the deepest conceptual point. In everyday life a feather drifts slowly and an apple drops fast because air pushes on the large surface of the feather. Remove the air, and the difference disappears.

Stroboscopic view of a green apple and white feather falling in vacuum side by side at same rate with constant acceleration g downward

A classic demonstration uses stroboscopic photos taken at equal time intervals. Both objects fall, both accelerate downward, both at the same rate \(g\). Their speeds increase at the same rate, and they fall together. In a vacuum chamber on the Moon, Apollo astronauts dropped a hammer and a feather and they hit the ground simultaneously — not because of trick, but because \(g\) does not depend on mass.

Callout Box:Why Mass Does Not Matter: In free-fall, Newton's Second Law and Newton's Law of Gravitation both contain mass. The gravitational force is larger for a heavier object, but its inertia is larger by the same factor. The two cancel, leaving \(a = g\) for all masses. Air resistance breaks this cancellation because drag depends on shape and area, not mass.

The value of \(g\) varies slightly with latitude and with elevation. It is slightly larger near the poles and slightly smaller near the equator because Earth is not a perfect sphere and rotates. It also decreases with altitude. At sea level in Earth's midlatitudes the value is \(9.8\,\text{m/s}^2\) \((32\,\text{ft/s}^2)\), which is what you should use as an exact number for the problems in this textbook unless otherwise noted.

Location	g value	Why it differs
Sea level, mid-latitude (standard)	\(9.80\,\text{m/s}^2\)	Reference value for exams
Equator, sea level	\(9.78\,\text{m/s}^2\)	Earth's rotation, larger radius
North Pole, sea level	\(9.83\,\text{m/s}^2\)	No centrifugal reduction, closer to center
400 km ISS orbit	\(8.7\,\text{m/s}^2\)	Still strong, weightlessness is free-fall

Exam Tip:For AP, A-Level, and IB, always use \(g = 9.8\,\text{m/s}^2\) or \(10\,\text{m/s}^2\) if the exam explicitly says to. Never use \(-9.8\) for \(g\) itself. \(g\) is a magnitude and is always positive. The acceleration is \(a = -g\).

The constant-acceleration equations apply to free fall near Earth's surface; that is, they apply to an object in vertical flight, either up or down, when the effects of air can be neglected. However, for free fall we apply two critical sign changes:

1. The directions of motion are now along a vertical \(y\) axis instead of a horizontal \(x\) axis, with the positive direction of \(y\) upward. This is important for later chapters when combined horizontal and vertical motions are examined, as in projectile motion.
2. The free-fall acceleration is negative — that is, downward on the \(y\) axis, toward Earth's center — and so it has the value \(-g\) in the equations.

Standard Horizontal Motion	Free-Fall Vertical Motion
Axis: \(x\), \(+x\) to the right	Axis: \(y\), \(+y\) upward
Acceleration: \(a\)	Acceleration: \(a = -g = -9.8\,\text{m/s}^2\)
Initial position: \(x_0\)	Initial position: \(y_0\)
Equations use \(x - x_0\)	Equations use \(y - y_0\)

Callout Box:The Sign Convention You Must Not Break: The free-fall acceleration near Earth's surface is \(a = -g = -9.8\,\text{m/s}^2\), and the magnitude of the acceleration is \(g = 9.8\,\text{m/s}^2\). Do not substitute \(-9.8\,\text{m/s}^2\) for \(g\). If a formula says \(y = v_0 t - \frac{1}{2}gt^2\), \(g\) is \(9.8\), the minus sign is already written for you.

Suppose you toss a tomato directly upward with an initial positive velocity \(v_0\) and then catch it when it returns to the release level. During its free-fall flight, from just after its release to just before it is caught, the constant-acceleration equations apply to its motion. The acceleration is always \(a = -g = -9.8\,\text{m/s}^2\), negative and thus downward. The velocity, however, changes.

During the ascent, the magnitude of the positive velocity decreases until it momentarily becomes zero. Because the tomato has then stopped, it is at its maximum height. During the descent, the magnitude of the now negative velocity increases.

Tomato tossed upward showing velocity decreasing positive to zero at top then increasing negative while acceleration stays -g constant downward

Phase	Sign of \(v\)	Sign of \(a\)	Speed Behavior
Ascent	\(v > 0\)	\(a = -g < 0\)	Speed decreasing, opposite signs
At top	\(v = 0\)	\(a = -g < 0\)	Momentarily stopped, still accelerating down
Descent	\(v < 0\)	\(a = -g < 0\)	Speed increasing, same signs

Exam Trap:At the highest point, many students say acceleration is zero because velocity is zero. It is not. Velocity is zero for an instant, but it is changing. The slope of \(v(t)\) at that instant is still \(-g\). If acceleration were zero at the top, the object would hang there.

Checkpoint Insight: If you toss a ball straight up, what is the sign of the ball's displacement for the ascent, from the release point to the highest point? Displacement \(y - y_0\) is positive because final position is above release. For the descent, from the highest point back to the release point, displacement from the top is negative, but displacement from the original release point back to release is zero at return. At its highest point, the ball's acceleration is still \(-g\), downward.

Complete Worked Analysis: Time for full up-down flight, baseball toss

A pitcher tosses a baseball up along a \(y\) axis, with an initial speed of \(12\,\text{m/s}\).

(a) How long does the ball take to reach its maximum height?

Key Ideas: (1) Once the ball leaves the pitcher and before it returns to his hand, its acceleration is the free-fall acceleration \(a = -g\). Because this is constant, the constant-acceleration equations apply. (2) The velocity \(v\) at the maximum height must be \(0\).

We know \(v\), \(a\), and initial velocity \(v_0 = 12\,\text{m/s}\), and seek \(t\), so we use \(v = v_0 + a t\).

\(\begin{aligned} t &= \frac{v - v_0}{a} \\ &= \frac{0 - 12\,\text{m/s}}{-9.8\,\text{m/s}^2} \\ &= 1.2\,\text{s} \end{aligned}\)

The time to rise is independent of mass and depends only on \(v_0/g\).

(b) What is the ball's maximum height above its release point?

We can take the ball's release point to be \(y_0 = 0\). We can then use \(v^2 = v_0^2 + 2a(y - y_0)\), set \(y - y_0 = y\) and \(v = 0\) at maximum height, and solve for \(y\).

\(\begin{aligned} y &= \frac{v^2 - v_0^2}{2a} \\ &= \frac{0 - (12\,\text{m/s})^2}{2(-9.8\,\text{m/s}^2)} \\ &= 7.3\,\text{m} \end{aligned}\)

Note this is about two stories high.

(c) How long does the ball take to reach a point \(5.0\,\text{m}\) above its release point?

We know \(v_0\), \(a = -g\), and displacement \(y - y_0 = 5.0\,\text{m}\), and want \(t\), so we choose \(y = y_0 + v_0 t + 1/2 a t^2\).

With \(y_0 = 0\):

\(y = v_0 t - 1/2 g t^2\)

\(5.0 = 12 t - 4.9 t^2\)

Rewriting as \(4.9t^2 - 12t + 5.0 = 0\).

Solving this quadratic equation for \(t\) yields \(t = 0.53\,\text{s}\) and \(t = 1.9\,\text{s}\).

There are two such times. This is not surprising because the ball passes twice through \(y = 5.0\,\text{m}\), once on the way up and once on the way down. The sum of the two times is \(2 \times 1.2 = 2.4\,\text{s}\), twice the time to the top, which is the total flight time back to \(y_0 = 0\) if it were caught at release level. The product corresponds to \(2y/g\).

Baseball tossed at 12 m/s showing y(t) parabola peaking at 7.3 m at 1.2 s and v(t) line slope -g crossing zero

Pro Tip for Exams:When you see "find the time when the object is at height \(y\)", expect a quadratic and two answers. The smaller is ascent, larger is descent. If the discriminant \(v_0^2 - 2gy\) is negative, the height is never reached.

Callout Box:Symmetry of Free-Fall: For a launch and return to the same level with no air: time up = time down, speed at a given height is same magnitude going up and down but opposite sign, and total flight time is \(2v_0/g\).

Practice Questions

MCQs (Multiple Choice Questions)

01. If air resistance is negligible, the free-fall acceleration of a 1 kg feather and a 10 kg hammer is:

1. Larger for hammer
2. Larger for feather
3. Same for both
4. Zero for both

Correct: C
\(g\) independent of mass, density, shape. Both accelerate at \(g\) downward.

02. In free-fall notation, \(+y\) is upward. The acceleration is:

1. \(+g\)
2. \(-g\)
3. \(0\) at top
4. \(+g\) on way up, \(-g\) on way down

Correct: B
Always \(a = -g = -9.8\,\text{m/s}^2\) throughout flight.

03. You toss a tomato up with \(v_0 = +10\,\text{m/s}\). At its maximum height:

1. \(v = 0\), \(a = 0\)
2. \(v = 0\), \(a = -g\)
3. \(v = +10\), \(a = -g\)
4. \(v\) negative, \(a\) zero

Correct: B
Velocity momentarily zero but still accelerating downward, otherwise it would stay.

04. \(g = 9.8\,\text{m/s}^2\) equals:

1. \(32\,\text{ft/s}^2\)
2. \(9.8\,\text{ft/s}^2\)
3. \(32\,\text{m/s}^2\)
4. \(1\,\text{m/s}^2\)

Correct: A
\(9.8\,\text{m/s}^2 = 32\,\text{ft/s}^2\) is the imperial equivalent given.

05. A ball is tossed up. Its displacement from release to max height is:

1. Negative
2. Positive
3. Zero
4. Cannot tell

Correct: B
Final y above initial, so \(y - y_0 > 0\).

06. For the baseball \(v_0 = 12\,\text{m/s}\), time to max height \(1.2\,\text{s}\), max height is about:

1. 5.0 m
2. 7.3 m
3. 12 m
4. 14.7 m

Correct: B
\(y = v_0^2 / (2g) = 144 / 19.6 = 7.3\,\text{m}\).

07. The ball is at \(y = 5.0\,\text{m}\) above release. How many times does this occur during full flight?

1. Once
2. Twice
3. Never
4. Three times

Correct: B
\(t = 0.53\,\text{s}\) up and \(1.9\,\text{s}\) down from quadratic.

08. During ascent, velocity is positive and acceleration negative. Speed is:

1. Increasing
2. Decreasing
3. Constant
4. Zero

Correct: B
Opposite signs → speed decreasing.

09. If you substitute \(g\) in formula \(y = v_0 t - 0.5 g t^2\), you should use:

1. \(-9.8\)
2. \(+9.8\)
3. \(0\)
4. \(-9.8\) on way up only

Correct: B
\(g\) is magnitude, \(+9.8\). Minus sign already in equation.

10. Value of \(g\) varies slightly with:

1. Mass of object
2. Color
3. Latitude and elevation
4. Velocity

Correct: C
\(g\) depends on distance from Earth's center and rotation, not object's properties.

FRQs (Free Response Questions)

FRQ 1: (AP 1 - Conceptual)
A student says heavy objects fall faster because \(g\) depends on mass. Explain why feather and apple fall together in vacuum using Newton's laws.

Rubric:

• Gravitational force proportional to mass \(F_g = mg\), inertia also proportional to mass, \(a = F/m = g\) cancels mass (2 pts)
• Air drag breaks this because drag ∝ area, not mass (1 pt)
• In vacuum, same \(a = -g\) (1 pt)
• Reference to experiment (1 pt)

FRQ 2: (A-Level - Calculation)
A ball thrown upward with \(v_0 = 15\,\text{m/s}\) from \(y_0 = 0\). (a) Time to max height, (b) max height, (c) velocity when \(y = 10\,\text{m}\) on way down.

Rubric:

• (a) \(t = v_0/g = 15/9.8 = 1.53\,\text{s}\) (1 pt)
• (b) \(y_{max} = v_0^2/2g = 11.5\,\text{m}\) (2 pts)
• (c) Use \(v^2 = v_0^2 -2g y\), \(v = -\sqrt{225 -196}= -5.4\,\text{m/s}\) negative for descent (2 pts)

FRQ 3: (IB HL - Graph)
Sketch \(y(t)\) and \(v(t)\) for tomato tossed up with \(v_0 = +10\,\text{m/s}\) returning to release. Mark \(v = 0\), \(a\) constant.

Rubric:

• \(y(t)\) parabola opening down, peak at \(t = v_0/g\) (1 pt)
• \(v(t)\) straight line slope \(-g\), intercept \(v_0\), zero at peak, negative after (2 pts)
• Mark \(a = -g\) constant, does not become zero at top (2 pts)

FRQ 4: (College - Quadratic)
Show that an object thrown up with \(v_0\) passes height \(h < v_0^2/2g\) twice and derive sum of times equals \(2v_0/g\).

Rubric:

• \(h = v_0 t - 0.5 g t^2\) → \(0.5 g t^2 - v_0 t + h = 0\) (1 pt)
• Discriminant \(v_0^2 -2gh > 0\) gives two real roots (1 pt)
• Sum of roots = \(2v_0/g\) by Vieta (2 pts)
• Physical meaning: symmetric flight (1 pt)

FRQ 5: (USAPhO - Variation)
\(g\) at ISS altitude 400 km is \(8.7\,\text{m/s}^2\), not zero. Why do astronauts appear weightless? Relate to free-fall.

Rubric:

• ISS and astronauts both in free-fall with \(a = -g_{local}\), continuously falling around Earth (2 pts)
• No normal force, sensation of weightlessness same as free-fall elevator (2 pts)
• Not due to zero \(g\), value still ∼90% of surface (1 pt)

Frequently Asked Questions

What is free-fall acceleration?

Free-fall acceleration is constant downward acceleration \(g\) experienced by any object moving only under gravity with negligible air resistance. Near Earth's surface its magnitude is \(9.8\,\text{m/s}^2\) and direction is toward Earth's center.

Is free-fall acceleration the same for all objects?

Yes, when air is negligible. Its magnitude \(g\) is independent of mass, density, or shape. A feather and an apple accelerate identically in vacuum, increasing speed at the same rate.

Why is acceleration -g in free-fall?

We choose \(+y\) upward. Free-fall acceleration points downward, so \(a = -g = -9.8\,\text{m/s}^2\). \(g\) itself is always positive \(9.8\,\text{m/s}^2\); the minus sign gives direction.

What is acceleration at the top of a throw?

Acceleration is still \(-g\) downward at the highest point. Velocity is zero there, but velocity is changing. If \(a\) were zero, the object would remain at rest.

How do you find max height in free-fall?

Use \(v^2 = v_0^2 + 2a(y-y_0)\) with \(v = 0\) at top and \(a = -g\). This gives \(y_{max}= y_0 + v_0^2/(2g)\). For \(v_0=12\,\text{m/s}\), \(y_{max}=7.3\,\text{m}\).

Why are there two times for same height?

Equation \(y = v_0 t - 0.5gt^2\) is quadratic in \(t\). A height below max is crossed once on ascent and once on descent, giving two solutions like \(0.53\,\text{s}\) and \(1.9\,\text{s}\) for \(5\,\text{m}\) with \(12\,\text{m/s}\) launch.

Does g change with location?

Slightly. \(g\) is about \(9.78\,\text{m/s}^2\) at equator, \(9.83\) at poles, and decreases with altitude. For exams use \(9.8\,\text{m/s}^2\) unless stated.

Can we use horizontal kinematic equations for free-fall?

Yes, same equations, just change \(x\) to \(y\) and \(a\) to \(-g\). \(v = v_0 - gt\), \(y = y_0 + v_0 t - 0.5gt^2\), \(v^2 = v_0^2 -2g(y-y_0)\).

Exam Mapping

Concept	A-Level Physics (AQA/Edexcel)	AP Physics 1 & C	IB Physics HL	College Physics 1 & 2	USAPhO / IPhO
\(g = 9.8\,\text{m/s}^2\) independent of mass	Required practical: free-fall \(g\), misconception about mass	Unit 1 Kinematics: \(g\) as known constant, no mass dependence	Topic 2.1: \(g\) determination, air resistance vs vacuum	Lab: \(g\) measurement, error analysis	Foundation for all gravitational problems
\(a = -g\), \(+y\) up	Vector sign convention, \(y = ut + 0.5 a t^2\) with \(a = -g\)	Sign convention MCQ traps, \(a\) always down	IA: sign errors in projectile motion	Kinematic equations with \(a = -g\)	Coordinate choice, sign consistency
\(v = 0\) at max height, \(a = -g\) still	Common exam question: velocity vs acceleration at top	Classic conceptual trap: \(a \neq 0\) at top	Conceptual: acceleration at apex	Conceptual check: top point	Critical for energy turning points
Two times for same \(y\)	Quadratic use in \(s = ut + 0.5 a t^2\)	Quadratic solution, symmetry	Paper 2: solving quadratic for time	Algebraic problem solving	Symmetry \(t_{up} + t_{down} = 2v_0/g\)
Full flight \(v_0 = 12\,\text{m/s}\) → \(t = 1.2\,\text{s}\), \(7.3\,\text{m}\)	Calculations of time to apogee and height	FRQ: show work with \(v = v_0 + at\)	Calculation of max height	Example problem in textbooks	Basis for projectile max height

Conclusion

Free-fall distills constant acceleration to its purest form: a single downward \(a = -g\) that never changes, never depends on mass, and acts while velocity goes from positive to zero to negative. You now know to switch from \(x\) to \(y\) with \(+y\) up, to keep \(g = 9.8\) as a magnitude and write \(a = -g\), to read ascent as speed decreasing with opposite signs and descent as speed increasing with same signs, and to expect two times for any height below the peak from \(4.9t^2 -12t +5.0 = 0\). With \(t_{up} = 1.2\,\text{s}\) and \(y_{max}=7.3\,\text{m}\) for \(12\,\text{m/s}\), you can reconstruct any throw. This is the launchpad for projectile motion where this same vertical free-fall occurs while horizontal velocity stays constant.

References

Fundamentals of Physics by Halliday, Resnick, and Walker, Latest Edition, Chapter 2.5 Free-Fall Acceleration - Primary source for free-fall definition, \(g = 9.8\,\text{m/s}^2 = 32\,\text{ft/s}^2\), feather and apple vacuum demonstration, notation change to \(y\) with \(+y\) up and \(a = -g\), tomato ascent-descent analysis, and sample problem baseball toss at \(12\,\text{m/s}\).