What is Free Fall Class 9? Notes, Formula & Examples

If you drop a tennis ball and a textbook from your desk, they both land on the floor at the exact same instant. But if you drop a feather and a stone, the stone crashes down while the feather drifts gently from side to side. Why does this happen? And what really happens when an object drops under gravity alone?

In physics, understanding what is free fall class 9 students encounter in their exams comes down to one single idea: gravity working completely alone without interference. In this comprehensive guide, we will break down the NCERT definition of free fall, explore acceleration due to gravity, modify the three equations of motion, solve exam-style numericals step by step, and see how you can test these concepts yourself.

What is Free Fall? (NCERT Class 9 Definition)

Whenever an object falls toward the Earth under the influence of gravitational force alone, we say that the object is in a state of free fall.

According to NCERT Class 9 Science — Gravitation, the defining condition of free fall is that no other force acts on the moving body except the Earth's gravitational attraction. In a typical classroom or outdoors, air resistance (the upward friction from air molecules) pushes against falling objects. Because of this, falling through ordinary air is not technically "pure" free fall.

True free fall occurs strictly in a vacuum—a space entirely empty of air or other gases. However, for relatively heavy, streamlined objects like stones, metal balls, or cricket balls falling short distances, air resistance is tiny enough that we can treat their motion as free fall in our school physics calculations.

During free fall, notice three key physical facts:

• Direction remains constant: The object moves straight toward the center of the Earth.
• Velocity increases continuously: Because Earth constantly pulls the object downward, its speed increases steadily every second.
• Acceleration is uniform: Since the velocity changes at a constant rate, the falling object experiences constant uniform acceleration.

Acceleration Due to Gravity (g)

When an object is in free fall, its velocity is constantly changing. A change in velocity over time means the object is accelerating. This acceleration produces a fundamental constant in introductory physics.

What is g and Why is it Constant Near Earth's Surface?

The uniform acceleration experienced by a freely falling body solely due to the gravitational pull of the Earth is called acceleration due to gravity. It is denoted by the lowercase letter g.

From Newton's second law of motion, force is mass times acceleration (F = m × a). For a falling body of mass m, the gravitational force acting on it is F = m × g. At the same time, according to Newton's universal law of gravitation, the force between the Earth (mass M) and the object is:

F = (G × M × m) / R²

Equating the two expressions for force:

m × g = (G × M × m) / R²

Dividing both sides by m gives:

g = (G × M) / R²

Notice something remarkable about this equation: the mass of the falling object (m) cancels out completely! This proves that near the Earth's surface, g does not depend on the mass, size, or shape of the falling body. A bowling ball and a raindrop experience the exact same acceleration due to Earth's gravity.

When you plug in the standard values for Earth:

• Universal Gravitational Constant: G = 6.67 × 10⁻¹¹ N m²/kg²
• Mass of Earth: M = 6 × 10²⁴ kg
• Radius of Earth: R = 6.4 × 10⁶ m

You get the standard value of acceleration due to gravity:

g = 9.8 m/s²

This means that for every second an object is freely falling downward, its downward velocity increases by 9.8 meters per second. For quick classroom calculations or when explicitly stated in NCERT exam problems, you may round this value to g = 10 m/s².

Difference Between Small 'g' and Capital 'G'

One of the most common mistakes in Class 9 physics exams is confusing small 'g' with capital 'G'. They represent two very different physical concepts:

Parameter	Acceleration Due to Gravity (g)	Universal Gravitational Constant (G)
Definition	Acceleration produced in a freely falling body due to gravity.	Gravitational force between two 1 kg masses placed 1 meter apart.
Nature of Quantity	Vector quantity (acts vertically downward toward Earth's center).	Scalar quantity (a fundamental physical constant).
Standard Value	9.8 m/s² near Earth's surface.	6.673 × 10⁻¹¹ N m²/kg².
SI Unit	m/s² (or N/kg)	N m²/kg²
Constancy	Varies from place to place (higher at poles, lower at equator, zero at Earth's center).	Remains constant throughout the entire universe.

Modified Equations of Motion for Free Fall

In kinematics, you learned the three standard equations of motion for objects moving with uniform acceleration a:

1. v = u + at
2. s = ut + ½at²
3. v² = u² + 2as

Where u is initial velocity, v is final velocity, t is time taken, a is acceleration, and s is displacement.

Because acceleration during free fall is constant (a = g) and displacement is vertical height (s = h), we can directly adapt these formulas for freely falling bodies.

The Free Fall Equations

When an object is dropped or moves downward under gravity:

1. v = u + gt
2. h = ut + ½gt²
3. v² = u² + 2gh

If an object is simply dropped from rest, its initial velocity is zero (u = 0). Under these conditions, the formulas become delightfully simple:

• v = gt
• h = ½gt²
• v² = 2gh (or v = √(2gh))

Sign Conventions Made Easy (Exam Tip)

Students often lose marks in numerical problems by mixing up plus and minus signs. To prevent errors, follow this consistent sign convention:

• Case 1: Downward Motion (Object Dropped or Thrown Down)
  • Direction of motion is downward.
  • Gravity acts downward, speeding up the object.
  • Take a = +g (+9.8 m/s²).
  • Height fallen (h) is positive.
  • If dropped from rest, set u = 0.
• Case 2: Upward Motion (Object Thrown Vertically Upward)
  • Direction of motion is upward.
  • Gravity acts downward, opposing the motion and slowing the object down.
  • Take a = -g (-9.8 m/s²).
  • At the highest point, the object stops momentarily before falling back: v = 0.
  • Time to reach the highest point equals the time taken to fall back to the release point (Time of ascent = Time of descent).

Interactive Lab: The Feather and Coin Experiment

Why did thinkers before modern science believe heavy objects fall faster than light ones? Because in daily life, if you drop a feather and a metal coin simultaneously, the coin strikes the ground in a fraction of a second while the feather hovers through the air.

The culprit is air resistance. The feather has a large surface area relative to its small mass, so the upward push of air quickly balances its tiny weight. The coin, being dense and compact, cuts through the air with negligible drag.

To investigate whether mass really changes falling speed, scientists designed the famous feather and coin experiment. A coin and a feather are placed inside a long glass tube connected to a vacuum pump:

• With air inside the tube: When inverted, the coin drops straight to the bottom, while the feather floats down slowly.
• With air pumped out (vacuum): When the air is completely removed and the tube is flipped, both the feather and the coin fall side-by-side and hit the bottom at the exact same instant!

In 1971, astronaut David Scott recreated this test on the Moon during the Apollo 15 mission. Because the Moon has no atmosphere, he dropped a 1.32 kg geological hammer and a 0.03 kg falcon feather simultaneously. They struck the lunar dust together, offering clear visual proof that in the absence of air drag, all objects accelerate at the exact same rate under gravity.

To understand the historical experiments that debunked earlier myths, read our detailed guide on how Galileo explained why heavy and light objects fall at the same rate. You can also explore these physics principles directly in the Visiora interactive lab, adjusting gravity, initial height, and atmospheric conditions in real time.

Try It at Home: The Two-Sheet Paper Experiment

You do not need a vacuum chamber to test this principle at home:

1. Take two identical sheets of paper from your notebook (both have the same mass).
2. Keep one sheet completely flat.
3. Crumple the second sheet into a tight, dense ball.
4. Hold both at shoulder height and release them at the exact same moment.

The crumpled paper hits the floor instantly, while the flat sheet flutters down slowly. Because both sheets have identical mass, this proves that shape and air resistance—not mass—caused the difference in fall time!

Mass, Weight, and Weightlessness in Free Fall

A fascinating consequence of free fall is the sensation of weightlessness.

What is weight? Scientifically, weight is the gravitational force exerted by the Earth on a body (W = m × g). But the sensation of weight you feel on your feet right now is actually the normal reaction force of the ground pushing back up against you.

Imagine standing on a weighing scale inside an elevator. If the elevator cables snap and the cabin plunges in free fall:

• Both you and the scale accelerate downward at g = 9.8 m/s².
• Your feet no longer press against the scale, and the scale does not push against your feet.
• The scale reads 0 kg.

Your actual mass has not changed, and the Earth is still pulling on you with gravitational force, but your apparent weight is zero. This condition is called weightlessness. Astronauts aboard the International Space Station (ISS) experience this exact same phenomenon. The ISS is not beyond Earth's gravity; it is in continuous free fall around the Earth, resulting in perpetual apparent weightlessness.

Solved Class 9 Free Fall Numericals (Step-by-Step)

Let's work through two classic NCERT numerical problems so you can master the steps required for full marks in your school exams.

Problem 1: Object Dropped from a Tower

Question: A stone is dropped from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground, and find the total time taken to reach the ground. (Take g = 9.8 m/s²)

Solution:

Step 1: Write down the given values and identify what to find.

• Initial velocity (u) = 0 m/s (since the stone is dropped from rest)
• Height / displacement (h) = 19.6 m
• Acceleration (a) = +g = +9.8 m/s² (motion is downward)
• Final velocity (v) = ?
• Time taken (t) = ?

Step 2: Calculate final velocity (v) using the third equation of motion.

v² = u² + 2gh

v² = 0² + 2 × 9.8 × 19.6

v² = 19.6 × 19.6

v = √(19.6²)

v = 19.6 m/s

Step 3: Calculate the time taken (t) using the first equation of motion.

v = u + gt

19.6 = 0 + 9.8 × t

t = 19.6 / 9.8

t = 2 s

Answer: The stone strikes the ground with a final velocity of 19.6 m/s after 2 seconds.

Problem 2: Object Thrown Vertically Upward

Question: A cricket ball is thrown vertically upward with a velocity of 29.4 m/s. Find: (a) the maximum height to which it rises, and (b) the total time it takes to return to the hands of the thrower. (Take g = 9.8 m/s²)

Solution:

Step 1: Identify given values for upward journey.

• Initial velocity (u) = 29.4 m/s
• Final velocity at the top (v) = 0 m/s
• Acceleration (a) = -g = -9.8 m/s² (motion opposes gravity)
• Maximum height (h) = ?
• Time of ascent (t₁) = ?

Step 2: Find maximum height (h).

v² = u² - 2gh

0 = (29.4)² - 2 × 9.8 × h

19.6 × h = (29.4)²

19.6 × h = 864.36

h = 864.36 / 19.6

h = 44.1 m

Step 3: Find time taken to reach the maximum height (t₁).

v = u - gt₁

0 = 29.4 - 9.8 × t₁

9.8 × t₁ = 29.4

t₁ = 29.4 / 9.8 = 3 s

Step 4: Find total time of flight.

Since the time taken to rise equals the time taken to fall back to the release point:

Total time of flight = 2 × t₁ = 2 × 3 s = 6 s

Answer: The ball rises to a maximum height of 44.1 m and returns after a total time of 6 seconds.

Frequently Asked Questions (Class 9 Exam Prep)

Is a skydiver with an open parachute in free fall?

No. Once the parachute deploys, significant upward drag from the air opposes gravitational pull. The skydiver reaches a steady terminal velocity where upward air drag balances downward weight. Because gravity is no longer the sole force acting, this is not true free fall.

Does the value of g depend on the mass of the falling object?

No. From the derivation g = GM / R², the value of g depends solely on the mass and radius of the Earth, not on the mass of the falling object. All bodies, light or heavy, accelerate at 9.8 m/s² near Earth's surface when air resistance is absent.

What is the acceleration of a ball at the highest point of its vertical path?

At the highest point, the instantaneous velocity of the ball is 0 m/s, but its acceleration is still 9.8 m/s² downward. Earth's gravity never turns off; it constantly acts toward the center of the Earth throughout the entire flight.

Why is g slightly greater at the poles than at the equator?

The Earth is an oblate spheroid—slightly flattened at the poles and bulging at the equator. This means the polar radius is slightly smaller than the equatorial radius. Because g is inversely proportional to R² (g ∝ 1 / R²), smaller radius at the poles produces a slightly higher value of g (around 9.83 m/s² at the poles versus 9.78 m/s² at the equator).

Summary: Key Takeaways for Class 9 Physics

• Free fall is the motion of an object falling solely under Earth's gravitational force.
• In pure free fall, acceleration is constant: g = 9.8 m/s².
• All objects accelerate at the same rate regardless of their mass when air drag is eliminated.
• Use a = +g for downward motion and a = -g for upward motion in kinematic equations.
• Objects experience weightlessness during free fall because apparent weight is zero.