Motion Under Gravity (Vertical Motion) NEET Notes with Solved Examples

Motion Under Gravity (Vertical Motion)

Easy NEET Notes

1. Situation in Figure

A ball is thrown vertically upward from the top of a building.

• Height of building = 25 m
• Initial velocity of ball = 20 m/s upward
• Acceleration due to gravity acts downward
• Take upward direction as positive

Therefore:

• Initial velocity → u = +20 m/s
• Acceleration → a = -10 m/s²

Gravity is always downward.

PART 1: Motion from A to B (Upward Motion)

At highest point B:

• Final velocity becomes zero
• Because the ball stops for a moment before coming down

Use first equation of motion:

v = u + at

Substitute values:

0 = 20 + (-10)t

0 = 20 - 10t

10t = 20

t = 2 s

Conclusion

t₁ = 2 s

Time taken to go upward = 2 seconds

PART 2: Motion from B to C (Falling Down)

Now the ball falls from highest point to ground.

At point B:

• Initial velocity = 0
• Acceleration = -10 m/s²

Maximum height above ground:

25 + 20 = 45 m

(25 m building + 20 m rise)

Use second equation of motion:

y = y₀ + v₀t + 1/2 at²

Substitute values:

0 = 45 + 0 + 1/2(-10)t²

0 = 45 - 5t²

5t² = 45

t² = 9

t = 3 s

Conclusion

t₂ = 3 s

Time to fall downward = 3 seconds

Total Time of Flight

T = t₁ + t₂

T = 2 + 3

T = 5 s

Final Answer

5 s

Total time taken by the ball to hit ground.

SECOND METHOD (Shortcut Method)

Use whole motion directly.

Given:

• Initial height = 25 m
• Final height = 0
• Initial velocity = 20 m/s
• Acceleration = -10 m/s²

Use:

y = y₀ + ut + 1/2 at²

Substitute:

0 = 25 + 20t + 1/2(-10)t²

0 = 25 + 20t - 5t²

5t² - 20t - 25 = 0

Solve quadratic equation:

t = 5 s

Same answer obtained.

Important NEET Concepts

1. Sign Convention

If upward is positive:

• Upward velocity → positive
• Downward velocity → negative
• Gravity → negative

2. At Maximum Height

v = 0

Velocity becomes zero only for an instant.

3. Acceleration Due to Gravity

Near Earth:

g = 9.8 m/s²

For NEET numericals often take:

g = 10 m/s²

Free Fall Notes (Example 2.4)

Definition of Free Fall

When an object moves only under gravity (air resistance neglected), it is called free fall.

Important Equations of Free Fall

Velocity-Time Relation

v = u - gt

If object released from rest:

u = 0

Then:

v = -gt

Position-Time Relation

y = y₀ + ut - 1/2 gt²

If released from rest and origin at starting point:

y = -1/2 gt²

Velocity-Position Relation

v² = u² - 2gy

If released from rest:

v² = -2gy

NEET Quick Tips

• Gravity always acts downward
• At highest point velocity becomes zero
• Acceleration never becomes zero
• Time going up ≠ time coming down if starting and ending heights are different

Most Important Formula Sheet

Equation	Formula
First Equation	v = u + at
Second Equation	s = ut + 1/2 at²
Third Equation	v² = u² + 2as

NEET Memory Trick

“UVS”

Remember:

• (u) → initial velocity
• (v) → final velocity
• (s) → displacement
• (a) → acceleration
• (t) → time

Use any equation depending on missing quantity.

NEET Physics Easy Notes © 2026

 Motion Under Gravity Class 11 Physics Notes for NEET & JEE 

Motion under gravity showing a ball thrown upward from a building with velocity, height, and time calculations.

CBSE Class 11 Physics: Motion Under Gravity Important Questions and Answers

These exam-oriented questions and answers cover MCQs, very short answer questions, short answer questions, long answer questions, assertion-reason questions, fill in the blanks, case study questions, statement-based questions, and match the columns.

Multiple Choice Questions (MCQs)

Question 1

A ball is thrown vertically upward with an initial velocity of 20 m/s. What is its velocity at the highest point?

Options:

• A) 20 m/s
• B) 10 m/s
• C) 0 m/s
• D) -20 m/s

Answer: C) 0 m/s

Question 2

What is the value of acceleration due to gravity near the Earth's surface?

Options:

• A) 0 m/s²
• B) 9.8 m/s² upward
• C) 9.8 m/s² downward
• D) 20 m/s² downward

Answer: C) 9.8 m/s² downward

Question 3

At the highest point of vertical motion, the acceleration of the body is:

Options:

• A) Zero
• B) 9.8 m/s² upward
• C) 9.8 m/s² downward
• D) Infinite

Answer: C) 9.8 m/s² downward

---

Very Short Answer Questions

Question 1

What is free fall?

Answer: The motion of an object under the influence of gravity alone is called free fall.

Question 2

What is the velocity of a body at the highest point of its motion?

Answer: The velocity becomes zero at the highest point.

Question 3

What is the SI unit of acceleration?

Answer: metre per second squared (m/s²).

Question 4

Does acceleration become zero at the highest point?

Answer: No. Acceleration remains equal to acceleration due to gravity and acts downward.

---

Short Answer Questions

Question 1

Why does a ball thrown upward slow down as it rises?

Answer: A ball thrown upward slows down because gravity acts downward, opposite to the direction of motion. This downward acceleration continuously decreases the velocity until it becomes zero at the highest point.

Question 2

Write the equations of motion under gravity.

Answer:

For upward positive direction:

v = u − gt

s = ut − ½gt²

v² = u² − 2gs

where:

• u = initial velocity
• v = final velocity
• g = acceleration due to gravity
• s = displacement
• t = time

Question 3

Differentiate between velocity and acceleration at the highest point.

Answer:

Velocity	Acceleration
Velocity becomes zero.	Acceleration remains 9.8 m/s² downward.
Changes direction after reaching the top.	Direction remains downward throughout the motion.

---

Long Answer Questions

Question 1

A ball is thrown vertically upward from the top of a building 25 m high with an initial velocity of 20 m/s. Calculate the total time taken to reach the ground. Take g = 10 m/s².

Answer:

For upward motion:

v = u − gt

0 = 20 − 10t

t = 2 s

Therefore, time to reach the highest point = 2 seconds.

Maximum height gained above the building:

h = u² / 2g

h = (20)² / (2 × 10)

h = 400 / 20

h = 20 m

Total height above ground:

25 + 20 = 45 m

For downward motion:

s = ½gt²

45 = 5t²

t² = 9

t = 3 s

Time of descent = 3 seconds.

Total time of flight:

T = 2 + 3

T = 5 s

Final Answer: The ball reaches the ground after 5 seconds.

---

Assertion and Reason Questions

Question 1

Assertion (A): Velocity becomes zero at the highest point.

Reason (R): Gravity stops acting at the highest point.

Answer: Assertion is true, but Reason is false because gravity acts continuously throughout the motion.

Question 2

Assertion (A): Acceleration due to gravity remains constant during vertical motion.

Reason (R): Gravity acts continuously downward.

Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

---

Fill in the Blanks

Question 1: The acceleration due to gravity always acts __________.

Answer: downward

Question 2: At the highest point of motion, velocity becomes __________.

Answer: zero

Question 3: Motion under gravity alone is called __________.

Answer: free fall

Question 4: The SI unit of acceleration is __________.

Answer: m/s²

---

Case Study Questions

A student throws a ball vertically upward from the top of a building with an initial velocity of 20 m/s. The height of the building is 25 m. Assume g = 10 m/s².

Question 1

How much time does the ball take to reach the highest point?

Answer: 2 seconds.

Question 2

What is the maximum height gained above the building?

Answer: 20 metres.

Question 3

What is the total height above the ground at the highest point?

Answer: 45 metres.

Question 4

What is the velocity at the highest point?

Answer: 0 m/s.

Question 5

What is the total time of flight?

Answer: 5 seconds.

---

Match the Columns

Column A	Column B
Highest Point	Velocity becomes zero
Free Fall	Motion under gravity alone
Acceleration Due to Gravity	9.8 m/s²
SI Unit of Acceleration	m/s²

Answers:

• Highest Point → Velocity becomes zero
• Free Fall → Motion under gravity alone
• Acceleration Due to Gravity → 9.8 m/s²
• SI Unit of Acceleration → m/s²

Important Exam Tip

Remember that at the highest point of vertical motion, velocity becomes zero but acceleration due to gravity remains constant and acts downward. This is one of the most frequently asked concepts in CBSE Class 11 Physics examinations. 

 Internal Links

Motion in a Straight Line Notes

Kinematics Complete Revision Notes

Difference Between Speed and Velocity

Free Fall Numerical Problems

Acceleration Due to Gravity Explained

Vector Basics for NEET

Projectile Motion Notes

NCERT Solutions – Motion in a Straight Line

Important Kinematics Formula Sheet

Previous Year NEET Kinematics Questions

Example 3.4 Solution Explained for Beginners

Block and Trolley System NEET Solution

Newton's Laws of Motion Notes

Work, Energy and Power Notes

Fundamental Forces in Nature Guide

Projectile Motion Example Solved Step by Step for Beginners

 -Dr.Sanjaykumar pawar

Step-by-step projectile motion example showing maximum height, time of flight, and horizontal range.

Internal Links

Laws of Motion Notes

Motion in Two Dimensions

Kinematics Formula Sheet

Newton’s Laws Numerical Problems

Velocity and Acceleration Explained

Circular Motion Notes

Physics Class 11 Important Questions

Projectile Motion Formula Derivation

Work Energy Theorem Notes

Gravitation Chapter Notes

Example 3.8 - Projectile Motion

A cricket ball is thrown at a speed of 28 m/s in a direction 30° above the horizontal.

Calculate:

1. Maximum Height
2. Time Taken to Return to Same Level
3. Horizontal Range

---

Step 1: Given Values

Initial velocity (u) = 28 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

---

(a) Maximum Height

Formula: H = (u² sin²θ) / 2g

H = (28² × sin²30°) / (2 × 9.8)

sin30° = 1/2

H = (28² × (1/2)²) / 19.6

28² = 784

(1/2)² = 1/4

H = (784 × 1/4) / 19.6

H = 196 / 19.6

H = 10 m

Answer: Maximum Height = 10 m

---

(b) Time of Flight

Formula: T = (2u sinθ) / g

T = (2 × 28 × sin30°) / 9.8

sin30° = 1/2

T = (2 × 28 × 1/2) / 9.8

T = 28 / 9.8

T = 2.86 s

Approximate value = 2.9 s

Answer: Time of Flight = 2.9 s

---

(c) Horizontal Range

Formula: R = (u² sin2θ) / g

R = (28² × sin(2 × 30°)) / 9.8

2 × 30° = 60°

R = (28² × sin60°) / 9.8

sin60° = 0.866

R = (784 × 0.866) / 9.8

R = 678.944 / 9.8

R ≈ 69 m

Answer: Horizontal Range = 69 m

---

Final Answers

Quantity	Answer
Maximum Height	10 m
Time of Flight	2.9 s
Horizontal Range	69 m

Projectile Motion Class 11 Physics Notes with Formulas for NEET

  Projectile Motion Formulas, Range, Height & Time of Flight – NEET

PROJECTILE MOTION

│

├── Definition

│   ├── Object projected at an angle

│   ├── Moves under gravity only

│   └── Path followed = Parabola

│

├── Components of Velocity

│   │

│   ├── Horizontal Component

│   │   ├── vx = v₀ cosθ₀

│   │   ├── Constant

│   │   └── No horizontal acceleration

│   │

│   └── Vertical Component

│       ├── vy = v₀ sinθ₀ − gt

│       ├── Changes continuously

│       └── Affected by gravity

│

├── Equation of Path

│   ├── y = x tanθ₀ − (g x²)/(2 v₀² cos²θ₀)

│   ├── Relation between x and y

│   └── Form = y = ax − bx²

│

├── Shape of Path

│   ├── Equation of parabola

│   └── Projectile path is parabolic

│

├── Maximum Height

│   │

│   ├── At top point

│   │   └── vy = 0

│   │

│   ├── Formula

│   │   └── hₘ = (v₀² sin²θ₀)/(2g)

│   │

│   └── Depends on

│       ├── Initial velocity

│       └── Angle of projection

│

├── Time to Reach Maximum Height

│   ├── tm = (v₀ sinθ₀)/g

│   └── Greater vertical velocity → greater time

│

├── Time of Flight

│   ├── Total time in air

│   ├── Tf = (2 v₀ sinθ₀)/g

│   └── Relation

│       └── Tf = 2tm

│

├── Horizontal Range

│   ├── Horizontal distance travelled

│   ├── R = (v₀² sin2θ₀)/g

│   └── Depends on

│       ├── Initial velocity

│       └── Angle of projection

│

├── Maximum Range

│   ├── Occurs at θ₀ = 45°

│   ├── sin2θ₀ = 1

│   └── Rmax = v₀²/g

│

├── Complementary Angles

│   ├── θ and (90° − θ)

│   ├── Same range

│   └── Examples

│       ├── 30° and 60°

│       └── 20° and 70°

│

└── Important NEET Points

    ├── Path is parabola

    ├── Horizontal velocity constant

    ├── Vertical velocity changes

    ├── vy = 0 at maximum height

    ├── Maximum range at 45°

    └── Time of flight = 2 × time to reach maximum height

Projectile motion showing parabolic trajectory, maximum height, horizontal range, and time of flight formulas for NEET Physics.

- Dr.Sanjaykumar pawar

INTERNAL LINKS

Laws of Motion Notes for NEET

Motion in a Straight Line Notes

Motion in a Plane Complete Notes

Kinematics Formula Sheet

Vectors for NEET Physics

Work, Energy and Power Notes

Circular Motion Notes

Important Physics Formulas for NEET

Gravitation Notes for Beginners

NCERT Physics Chapter Wise Notes

Projectile Motion Notes (NEET Level)

1. Equation of Path of a Projectile

When an object is projected with initial velocity at an angle, it moves in both:

• Horizontal direction (x-direction)
• Vertical direction (y-direction)

The equation of path gives the relation between horizontal displacement and vertical displacement.

y = x tanθ₀ − (g x²) / (2 v₀² cos²θ₀)

Meaning of Symbols

• y = vertical displacement
• x = horizontal displacement
• v₀ = initial velocity
• θ₀ = angle of projection
• g = acceleration due to gravity

2. Shape of the Path

The equation is of the form:

y = ax − bx²

This is the equation of a parabola.

Therefore: The path followed by a projectile is always a parabola.

Important NEET Point

• Horizontal motion → uniform velocity
• Vertical motion → accelerated motion due to gravity

Combination of these motions produces a parabolic path.

3. Time to Reach Maximum Height

At maximum height, vertical velocity becomes zero.

vᵧ = v₀ sinθ₀ − gt

At maximum height:

vᵧ = 0

Therefore:

tₘ = (v₀ sinθ₀) / g

• tₘ = time to reach maximum height

Greater the vertical component of velocity, greater the time taken to reach maximum height.

4. Time of Flight

The total time during which the projectile remains in air is called Time of Flight.

T_f = (2 v₀ sinθ₀) / g

Important Relation

T_f = 2 tₘ

Projectile takes equal time to go upward and downward because of symmetry.

5. Maximum Height of Projectile

The maximum vertical distance reached by projectile is called maximum height.

hₘ = (v₀² sin²θ₀) / (2g)

Important Points

• Maximum height depends on initial velocity.
• Maximum height depends on angle of projection.
• Greater vertical velocity gives greater height.

6. Horizontal Range of Projectile

Horizontal distance travelled before touching the ground is called horizontal range.

R = (v₀² sin2θ₀) / g

• R = horizontal range

7. Condition for Maximum Range

Range becomes maximum when:

sin2θ₀ = 1

Therefore:

θ₀ = 45°

Important NEET Result: Maximum range occurs at angle 45°.

Maximum Range Formula

R_max = v₀² / g

8. Important NEET Tricks

Complementary Angles

Angles θ and (90° − θ) give the same range.

Examples:

• 30° and 60°
• 20° and 70°

Horizontal Velocity

v_x = v₀ cosθ₀

Horizontal velocity remains constant because no horizontal acceleration acts.

Vertical Velocity

v_y = v₀ sinθ₀ − gt

Vertical velocity changes continuously due to gravity.

9. Quick Formula Revision Table

Quantity	Formula
Equation of Path	y = x tanθ₀ − (g x²)/(2 v₀² cos²θ₀)
Time to Maximum Height	tₘ = (v₀ sinθ₀)/g
Time of Flight	T_f = (2 v₀ sinθ₀)/g
Maximum Height	hₘ = (v₀² sin²θ₀)/(2g)
Horizontal Range	R = (v₀² sin2θ₀)/g
Maximum Range	R_max = v₀²/g

10. One-Line NEET Revision

• Projectile path is a parabola.
• At maximum height, vertical velocity becomes zero.
• Time of flight is double the time to reach maximum height.
• Maximum range occurs at 45°.
• Complementary angles give same range.

Prepared for NEET Beginners Physics Revision

Projectile Motion Question Bank
CBSE Class 11 Physics

1. Multiple Choice Questions (MCQs)

Q1. The path followed by a projectile is:
a) Straight line
b) Circle
c) Parabola
d) Ellipse

Answer: c) Parabola

Q2. The horizontal velocity of a projectile:
a) Increases continuously
b) Decreases continuously
c) Remains constant
d) Becomes zero

Answer: c) Remains constant

Q3. At maximum height, vertical velocity becomes:
a) Maximum
b) Zero
c) Infinite
d) Constant

Answer: b) Zero

Q4. Maximum range occurs at angle:
a) 30°
b) 45°
c) 60°
d) 90°

Answer: b) 45°

2. Very Short Answer Questions

Q1. Define projectile motion.

Projectile motion is the motion of an object projected into air under the effect of gravity.

Q2. What is the shape of projectile path?

The shape of projectile path is parabola.

Q3. What is the acceleration acting on projectile?

Acceleration due to gravity (g).

Q4. At which point does vertical velocity become zero?

At maximum height.

3. Short Answer Questions

Q1. Why is projectile path parabolic?

Horizontal motion is uniform while vertical motion is accelerated due to gravity. Combining both motions produces a parabolic path.

Q2. Define horizontal range.

The horizontal distance travelled by projectile before reaching the ground is called horizontal range.

Q3. What are complementary angles?

Two angles whose sum is 90° are called complementary angles. Example: 30° and 60°.

Q4. Why does horizontal velocity remain constant?

Because no horizontal force acts on the projectile.

4. Long Answer Questions

Q1. Derive expression for horizontal range of projectile.

Horizontal range:

R = Horizontal velocity × Time of flight

Horizontal velocity:
v_x = v₀ cosθ

Time of flight:
T = (2v₀ sinθ)/g

Therefore,
R = v₀ cosθ × (2v₀ sinθ)/g

R = (2v₀² sinθ cosθ)/g

Using:
2 sinθ cosθ = sin2θ

Therefore,
R = (v₀² sin2θ)/g

Q2. Derive equation of trajectory of projectile.

Horizontal motion:

x = v₀ cosθ × t

t = x / (v₀ cosθ)

Vertical motion:

y = v₀ sinθ × t − ½gt²

Substituting value of t:

y = x tanθ − (gx²)/(2v₀² cos²θ)

This is the equation of trajectory.

5. Assertion and Reason Questions

Q1.

Assertion (A): Projectile path is parabolic.
Reason (R): Horizontal motion is uniform and vertical motion is accelerated.

Answer: Both A and R are true and R is correct explanation of A.

Q2.

Assertion (A): Range is maximum at 45°.
Reason (R): sin90° = 1.

Answer: Both A and R are true and R is correct explanation of A.

6. Fill in the Blanks

1. The path of projectile is a __________.

parabola

2. At maximum height, vertical velocity becomes __________.

zero

3. Maximum range occurs at angle __________.

45°

4. Horizontal velocity remains __________ during motion.

constant

7. Statement Based Questions

1. Projectile motion is two-dimensional motion.

True

2. Horizontal acceleration of projectile is zero.

True

3. Vertical velocity increases upward.

False

8. Match the Columns

Column A	Column B
1. Maximum range	a. Parabola
2. Shape of path	b. vy = 0
3. Maximum height	c. 45°
4. Horizontal motion	d. Uniform velocity

1 → c
2 → a
3 → b
4 → d

9. Case Study Questions

A boy throws a ball with velocity 20 m/s at an angle of 45°. The ball follows a curved path and returns to ground.

Q1. What type of motion is shown by the ball?

Q2. What is the shape of path?

Q3. At what angle is range maximum?

Q4. What happens to vertical velocity at highest point?

Q5. Which component of velocity remains constant?

1. Projectile motion

2. Parabola

3. 45°

4. Vertical velocity becomes zero

5. Horizontal component

10. Numerical Problem

A projectile is thrown with velocity 20 m/s at angle 30°. Find time of flight.

Given:

v₀ = 20 m/s
θ = 30°
g = 9.8 m/s²

Formula:
T = (2v₀ sinθ)/g

T = (2 × 20 × 0.5)/9.8

T = 2.04 s

Answer: 2.04 s

11. Important One-Line Questions

Q1. Which force acts on projectile after projection?

Gravitational force

Q2. What is the horizontal acceleration of projectile?

Zero

Q3. What is the SI unit of acceleration?

m/s²

Prepared for CBSE Class 11 Physics Revision