Newton's Second Law Notes for Class 11 Physics | CBSE & NEET Guide

 Newton's Second Law Explained: Projectile Motion and Force Components 

Easy Notes (NEET Level) – Important Points from Newton’s Second Law 

- Dr.Sanjaykumar pawar

1. Component of Velocity Normal (Perpendicular) to the Force Remains Unchanged

• A force can only change the velocity component in its own direction.
• The velocity component perpendicular (normal) to the force does not change.
• This is because there is no acceleration in the perpendicular direction.

Example: Projectile Motion

• In projectile motion, only gravitational force acts on the particle.
• Gravity acts vertically downward.
• Therefore, acceleration is only in the vertical direction.
• There is no horizontal force acting on the projectile (neglecting air resistance).
• Hence, the horizontal component of velocity remains constant throughout the motion.
• Only the vertical component of velocity changes due to gravity.

Key NEET Point:
Force changes velocity only along its direction; perpendicular velocity remains unchanged.

---

2. Newton's Second Law for a Single Particle

• Newton's second law is written as:

• Here, F is the net external force acting on the particle.
• a is the acceleration produced in the particle.
• This form is directly applicable to a single point particle.

Key NEET Point:
Always use the net external force, not individual forces separately.

---

3. Newton's Second Law for a System of Particles

• The same law can also be applied to:

  • A rigid body
  • A group of particles (system)

• In such cases:

  • F = total external force on the system.
  • a = acceleration of the centre of mass of the system.

• Internal forces between particles of the system are not included.

• Internal forces cancel each other and do not affect the motion of the whole system.

Example

• Two blocks connected by a string.
• Tension between the blocks is an internal force.
• Only external forces like gravity or an applied pull are considered.

Key NEET Point:
For a system, consider only external forces; ignore internal forces.

---

4. Acceleration Depends on Present Force Only

• Acceleration at any instant is determined by the force acting at that same instant.
• Past motion does not affect present acceleration.
• A particle has no "memory" of its previous motion.

Example: Stone Dropped from an Accelerated Train

• A train is moving with acceleration.

• A stone is dropped from the train.

• Immediately after release:

  • The stone is no longer connected to the train.
  • No horizontal force acts on the stone (ignoring air resistance).
  • Therefore, horizontal acceleration becomes zero.

• The stone keeps its horizontal velocity but does not keep the train's acceleration.

Key NEET Point:
Velocity may continue, but acceleration changes instantly according to the force acting at that moment.

---

5. Newton's Second Law is a Local Relation

• Newton's second law is called a local relation.
• It relates force and acceleration at the same place and same time.

Meaning

• Force acting here and now determines acceleration here and now.
• Previous positions, velocities, or accelerations do not directly determine present acceleration.
• Only the current force matters.

Key NEET Point:
Present acceleration depends only on present force, not on the history of motion.

---

Quick Revision for NEET

1. Force changes velocity only in its own direction.
2. Velocity perpendicular to force remains unchanged.
3. In projectile motion, horizontal velocity remains constant.
4. Newton's second law: .
5. For a system, use total external force only.
6. Internal forces are ignored.
7. Acceleration depends on the force acting at that instant.
8. A body has no memory of past acceleration.
9. Newton's second law is a local relation.
10. Present force determines present acceleration.  

Newton's Second Law explains how force produces acceleration and why horizontal velocity remains constant in projectile motion.

CBSE Class 11 Physics – Newton's Second Law (Important Questions with Answers)

A. MCQs (1 Mark Each)

1. In projectile motion, the horizontal component of velocity remains constant because:

a) Gravity acts horizontally
b) No horizontal force acts on the projectile
c) Air resistance is maximum
d) Vertical velocity is constant

Answer: (b) No horizontal force acts on the projectile

---

2. The force in Newton's second law represents:

a) Internal force
b) External force
c) Gravitational force only
d) Friction only

Answer: (b) External force

---

3. For a system of particles, acceleration refers to the acceleration of:

a) Any particle
b) Largest particle
c) Centre of mass
d) Geometric centre

Answer: (c) Centre of mass

---

4. Internal forces are:

a) Included in net external force
b) Ignored while applying Newton's second law to a system
c) Greater than external forces
d) Always zero

Answer: (b)

---

5. A body remembers:

a) Past acceleration
b) Past force
c) Present force only affects acceleration
d) Future force

Answer: (c)

---

B. Very Short Answer Questions (1 Mark)

1. What is meant by the net force on a particle?

Answer: The vector sum of all external forces acting on the particle.

---

2. Which velocity component remains unchanged in projectile motion?

Answer: Horizontal component of velocity.

---

3. What is the SI unit of force?

Answer: Newton (N).

---

4. Which force is not included while applying Newton's second law to a system?

Answer: Internal force.

---

5. What is meant by a local relation?

Answer: A relation that connects physical quantities at the same place and same instant.

---

C. Short Answer Questions (2–3 Marks)

1. Why does the horizontal velocity of a projectile remain constant?

Answer:

• Gravity acts vertically downward.
• No horizontal force acts on the projectile.
• Therefore, horizontal acceleration is zero.
• Hence, horizontal velocity remains constant.

---

2. What is the role of internal forces in a system?

Answer:

• Internal forces act between particles of the same system.
• They cancel each other in pairs.
• Therefore, they do not affect the motion of the system as a whole.

---

3. Explain why a stone dropped from an accelerating train has no horizontal acceleration.

Answer:

• After release, the stone is no longer connected to the train.
• No horizontal force acts on it.
• According to Newton's second law, acceleration depends on force.
• Hence horizontal acceleration becomes zero.

---

D. Long Answer Questions (3–5 Marks)

1. Explain Newton's second law for a system of particles.

Answer: Newton's second law states that the net external force acting on a body equals the product of its mass and acceleration.

For a system of particles:

• F represents the total external force on the system.
• a represents the acceleration of the centre of mass.
• Internal forces are not included because they cancel each other.
• The motion of the entire system depends only on external forces.

---

2. Explain why Newton's second law is called a local relation.

Answer:

• Force and acceleration are related at the same place and same instant.
• Present acceleration depends only on present force.
• Past motion does not influence present acceleration directly.
• A body has no memory of previous forces or accelerations.
• Therefore Newton's second law is called a local relation.

---

E. Assertion and Reason Questions

1.

Assertion (A): Horizontal velocity remains constant in projectile motion.

Reason (R): No horizontal force acts on the projectile.

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

---

2.

Assertion (A): Internal forces are included in the net force acting on a system.

Reason (R): Internal forces cancel each other.

Answer: Assertion is false, Reason is true.

---

3.

Assertion (A): A body remembers its previous acceleration.

Reason (R): Present acceleration depends only on present force.

Answer: Assertion is false, Reason is true.

---

F. Fill in the Blanks

1. In projectile motion, the ________ component of velocity remains constant. Answer: horizontal

2. Newton's second law relates force and ________. Answer: acceleration

3. For a system, acceleration refers to the acceleration of the ________. Answer: centre of mass

4. Internal forces are ________ while applying Newton's second law to a system. Answer: ignored

5. Present acceleration depends on ________ force. Answer: present

---

G. True or False

1. Gravity changes the horizontal velocity of a projectile. Answer: False

2. Internal forces affect the motion of the centre of mass. Answer: False

3. External force determines acceleration. Answer: True

4. Newton's second law is a local relation. Answer: True

5. A body remembers its past acceleration. Answer: False

---

H. Match the Columns

Column A	Column B
1. Projectile motion	(a) Centre of mass
2. System acceleration	(b) Horizontal velocity constant
3. Internal forces	(c) Present force
4. Local relation	(d) Ignored
5. Acceleration depends on	(e) Same place and time

Answers

1 → (b)
2 → (a)
3 → (d)
4 → (e)
5 → (c)

---

I. Statement-Based Questions

Statement 1:

A projectile is moving in air with negligible air resistance.

Statement 2:

Its horizontal velocity remains constant.

a) Both statements are true and Statement 2 explains Statement 1.
b) Both statements are true but Statement 2 does not explain Statement 1.
c) Statement 1 is true, Statement 2 is false.
d) Statement 1 is false, Statement 2 is true.

Answer: (a)

---

J. Case Study Questions (4 Marks)

Case Study

A train is moving with acceleration. A stone is dropped from the train. Immediately after release, the stone continues moving forward but experiences only gravitational force.

Questions

1. Which force acts on the stone after release?

Answer: Gravitational force.

2. What is the horizontal acceleration of the stone?

Answer: Zero.

3. Why does the stone continue moving horizontally?

Answer: Due to its existing horizontal velocity.

4. What does this example prove about Newton's second law?

Answer: Present acceleration depends only on present force and not on past motion.

---

Exam-Oriented One-Line Revision

• Force changes velocity only in its own direction.
• Velocity perpendicular to force remains unchanged.
• Horizontal velocity remains constant in projectile motion.
• Only external forces are considered for a system.
• Internal forces cancel each other.
• Acceleration of a system is acceleration of its centre of mass.
• Present force determines present acceleration.
• Newton's second law is a local relation.

NEWTON'S SECOND LAW – IMPORTANT POINTS

│

├── 1. Velocity Component Perpendicular to Force

│   │

│   ├── Force changes velocity only along its direction

│   ├── No acceleration perpendicular to force

│   ├── Perpendicular (normal) velocity remains constant

│   │

│   └── Example: Projectile Motion

│       │

│       ├── Gravity acts vertically downward

│       ├── Vertical velocity changes

│       ├── No horizontal force (air resistance neglected)

│       └── Horizontal velocity remains constant

│

├── 2. Newton's Second Law for a Particle

│   │

│   ├── F = ma

│   ├── F = Net external force

│   ├── a = Acceleration produced

│   └── Applicable to a single particle

│

├── 3. Newton's Second Law for a System

│   │

│   ├── Applicable to

│   │   ├── Rigid body

│   │   └── System of particles

│   │

│   ├── F = Total external force

│   ├── a = Acceleration of centre of mass

│   │

│   └── Internal Forces

│       │

│       ├── Not included in F

│       ├── Cancel each other

│       └── Do not affect system motion

│

├── 4. Acceleration Depends on Present Force

│   │

│   ├── Present force → Present acceleration

│   ├── Past motion does not matter

│   └── Body has no memory of past acceleration

│

├── 5. Example: Stone Dropped from Accelerated Train

│   │

│   ├── Train is accelerating

│   ├── Stone is released

│   ├── Connection with train ends

│   ├── No horizontal force on stone

│   ├── Horizontal acceleration = 0

│   └── Horizontal velocity continues unchanged

│

├── 6. Local Nature of Newton's Second Law

│   │

│   ├── Local Relation

│   │

│   ├── Force at a point

│   │       ↓

│   ├── Determines acceleration

│   │       ↓

│   └── At the same place and same time

│

└── NEET QUICK FACTS

    │

    ├── Force changes velocity only in its direction

    ├── Perpendicular velocity remains unchanged

    ├── Horizontal velocity is constant in projectile motion

    ├── F = ma

    ├── Use only external forces for a system

    ├── Ignore internal forces

    ├── Present force determines present acceleration

    ├── No memory of previous acceleration

    └── Newton's second law is a local relation 

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

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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

Example 3.7 Solution: Stone Thrown Horizontally from Cliff

  Horizontal Projectile Motion Problem Solved Step by Step

Dr.Sanjaykumar pawar

A stone thrown horizontally from a 490 m cliff follows projectile motion under gravity.

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Motion Under Gravity Examples

Example 3.7 – Horizontal Projection Motion

Example 3.7 A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 m s-1. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. (Take g = 9.8 m s-2 ).

A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 m s^-1. Neglecting air resistance, find:

• Time taken by the stone to reach the ground
• Speed with which it hits the ground

Take g = 9.8 m s^-2

---

Step 1: Write the Given Data

Height of cliff, h = 490 m

Initial horizontal velocity, u_x = 15 m s^-1

Initial vertical velocity, u_y = 0

Acceleration due to gravity, g = 9.8 m s^-2

Step 2: Find Time Taken to Reach the Ground

Use the equation of motion:

s = ut + ½gt²

Substitute the values:

490 = 0 × t + ½ × 9.8 × t²

490 = 4.9t²

t² = 490 / 4.9

t² = 100

t = √100

t = 10 s

Time taken to reach the ground = 10 s

Step 3: Find Vertical Velocity at Impact

Use the equation:

v = u + gt

Substitute the values:

v_y = 0 + (9.8 × 10)

v_y = 98 m s^-1

Step 4: Horizontal Velocity Remains Constant

v_x = 15 m s^-1

Step 5: Find Resultant Speed at Impact

Use Pythagoras theorem:

v = √(v_x² + v_y²)

Substitute the values:

v = √(15² + 98²)

v = √(225 + 9604)

v = √9829

v ≈ 99.1 m s^-1

Speed with which the stone hits the ground = 99.1 m s^-1

---

Final Answers

1. Time taken = 10 s

2. Speed at impact = 99.1 m s^-1

Galileo Projectile Motion Theorem Explained Step by Step

 Dr.sanjaykumar pawar 

Projectile motion diagram showing why complementary angles produce equal horizontal ranges. 

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Example 3.6 – Galileo’s Statement

Example 3.6 Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”. Prove this statement.

Question

Galileo stated:

“For elevations which exceed or fall short of 45° by equal amounts, the ranges are equal.”

Prove this statement.

---

Step 1: Write the Formula for Range

For a projectile projected with speed v₀ at an angle θ, the horizontal range is:

R = (v₀² sin 2θ) / g

Where:

• R = Horizontal Range
• v₀ = Initial Velocity
• g = Acceleration due to gravity
• θ = Angle of projection

---

Step 2: Consider Two Angles

Take two angles:

(45° + α)

and

(45° − α)

These angles are equally above and below 45°.

---

Step 3: Find 2θ for Both Angles

For the first angle:

2θ = 2(45° + α)
= 90° + 2α

For the second angle:

2θ = 2(45° − α)
= 90° − 2α

---

Step 4: Write the Sine Terms

The range depends on sin 2θ.

So we get:

sin(90° + 2α)

and

sin(90° − 2α)

Using trigonometric identities:

sin(90° + x) = cos x

sin(90° − x) = cos x

Therefore:

sin(90° + 2α) = cos 2α

sin(90° − 2α) = cos 2α

Both values are equal.

---

Step 5: Compare the Ranges

Since the value of sin 2θ is the same for both angles, the range formula gives the same result.

R₁ = R₂

---

Final Conclusion

Hence, the ranges are equal for projection angles that exceed or fall short of 45° by the same amount.

Therefore, Galileo’s statement is proved.

---

Quick Revision Point

Examples of equal ranges:

• 30° and 60°
• 40° and 50°

Because complementary angles produce equal horizontal ranges.

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

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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

Projectile Motion Class 11 Physics Notes for NEET Beginners

  Easy Projectile Motion Notes with Formulas for NEET Students

PROJECTILE MOTION

│

├── Definition

│   ├── Object thrown in air

│   ├── Moves under gravity only

│   └── Called projectile

│

├── Examples

│   ├── Cricket ball

│   ├── Football

│   ├── Stone

│   └── Bullet

│

├── Types of Motion

│   │

│   ├── Horizontal Motion

│   │   ├── Along x-axis

│   │   ├── No acceleration

│   │   ├── Constant velocity

│   │   └── Uniform motion

│   │

│   └── Vertical Motion

│       ├── Along y-axis

│       ├── Gravity acts downward

│       ├── Acceleration = g

│       └── Non-uniform motion

│

├── Galileo’s Concept

│   ├── Horizontal and vertical motions independent

│   └── Explained in 1632

│

├── Assumptions

│   ├── Air resistance neglected

│   └── Only gravity acts

│

├── Initial Velocity

│   ├── Total velocity = v₀

│   ├── Angle of projection = θ

│   │

│   ├── Horizontal Component

│   │   └── v₀x = v₀ cosθ

│   │

│   └── Vertical Component

│       └── v₀y = v₀ sinθ

│

├── Acceleration

│   ├── ax = 0

│   └── ay = -g

│

├── Position Equations

│   │

│   ├── Horizontal Position

│   │   └── x = (v₀ cosθ)t

│   │

│   └── Vertical Position

│       └── y = (v₀ sinθ)t - ½gt²

│

├── Velocity Equations

│   │

│   ├── Horizontal Velocity

│   │   └── vx = v₀ cosθ

│   │

│   └── Vertical Velocity

│       └── vy = v₀ sinθ - gt

│

├── Maximum Height

│   ├── Highest point

│   ├── vy = 0

│   └── Projectile changes direction

│

├── Trajectory

│   ├── Path followed by projectile

│   └── Shape = Parabola

│

├── Important NEET Points

│   ├── Horizontal velocity constant

│   ├── Gravity acts downward only

│   ├── vy = 0 at top point

│   ├── Path is parabolic

│   └── Motions are independent

│

└── Quick Tricks

    ├── cosθ → Horizontal

    ├── sinθ → Vertical

    ├── x-motion → Uniform

    ├── y-motion → Accelerated

    └── Gravity acts vertically downward

Projectile motion showing horizontal and vertical components of motion with parabolic trajectory.

Dr.Sanjaykumar pawar

Internal Links

Laws of Motion Notes for NEET

Motion in a Straight Line Notes

Motion in a Plane Complete Guide

Kinematics Formula Sheet

Gravitation Notes for NEET

Vectors Physics Notes

Work, Energy and Power Notes

Circular Motion NEET Notes

Physics Formula Revision Notes

NEET Physics Chapterwise Notes

Projectile Motion Notes for NEET Beginners

1. What is Projectile Motion?

A body thrown into the air and moving under the effect of gravity only is called a projectile.

The motion of such a body is called projectile motion.

Examples:

• Cricket ball
• Football
• Stone thrown in air
• Bullet fired at an angle

2. Main Idea of Projectile Motion

Projectile motion consists of two independent motions happening together.

(a) Horizontal Motion

• Motion along x-axis
• No acceleration acts horizontally
• Velocity remains constant

(b) Vertical Motion

• Motion along y-axis
• Gravity acts downward
• Acceleration is constant

3. Galileo’s Contribution

Galileo first explained that horizontal and vertical motions are independent of each other.

4. Assumption in Projectile Motion

• Air resistance is neglected.
• Only gravity acts on the projectile.

5. Initial Velocity of Projectile

Suppose a projectile is thrown with:

• Initial velocity = v₀
• Angle of projection = θ

6. Components of Initial Velocity

(a) Horizontal Component

v_0x = v₀ cos θ

• Acts along x-axis
• Remains constant throughout motion

(b) Vertical Component

v_0y = v₀ sin θ

• Acts along y-axis
• Changes due to gravity

7. Acceleration of Projectile

a_x = 0

a_y = -g

• No horizontal acceleration
• Gravity acts vertically downward

8. Initial Position

x₀ = 0

y₀ = 0

This means the projectile starts from the origin.

9. Position of Projectile at Time t

(a) Horizontal Position

x = (v₀ cos θ)t

• Horizontal distance increases uniformly
• Depends on time and initial velocity

(b) Vertical Position

y = (v₀ sin θ)t - ½gt²

• Gravity slows upward motion
• Gravity increases downward motion

10. Velocity Components at Any Time

(a) Horizontal Velocity

v_x = v₀ cos θ

Horizontal velocity remains constant.

(b) Vertical Velocity

v_y = v₀ sin θ - gt

Vertical velocity changes continuously because of gravity.

11. Maximum Height

At the highest point:

v_y = 0

• Projectile stops moving upward for a moment
• Then it starts moving downward

12. Shape of Projectile Path

The path followed by a projectile is called a trajectory.

The trajectory of projectile motion is always a parabola.

13. Important NEET Points

• Horizontal velocity remains constant.
• Gravity acts only downward.
• Vertical velocity becomes zero at maximum height.
• Projectile path is parabolic.
• Horizontal and vertical motions are independent.

14. Formula Summary

Horizontal Velocity:

v_x = v₀ cos θ

Vertical Velocity:

v_y = v₀ sin θ - gt

Horizontal Position:

x = (v₀ cos θ)t

Vertical Position:

y = (v₀ sin θ)t - ½gt²

15. Quick Revision Tricks

• cos θ → Horizontal component
• sin θ → Vertical component
• Horizontal motion → Constant velocity
• Vertical motion → Accelerated motion
• At highest point → v_y = 0

16. Conclusion

Projectile motion is a combination of:

• Uniform horizontal motion
• Vertically accelerated motion under gravity

Prepared for NEET Physics Beginners

Projectile Motion Question Bank

Class 11 CBSE Physics

1. Multiple Choice Questions (MCQs)

Q1. A projectile moves in a parabolic path because:

a) Horizontal velocity changes
b) Vertical velocity remains constant
c) Horizontal and vertical motions are independent
d) Gravity acts horizontally

Answer: c) Horizontal and vertical motions are independent

Q2. The acceleration of a projectile at highest point is:

a) Zero
b) g upward
c) g downward
d) Infinite

Answer: c) g downward

Q3. At maximum height, vertical velocity becomes:

a) Maximum
b) Minimum
c) Zero
d) Infinite

Answer: c) Zero

Q4. The horizontal component of velocity:

a) Increases
b) Decreases
c) Remains constant
d) Becomes zero

Answer: c) Remains constant

2. Very Short Answer Questions

Q1. What is a projectile?

A body thrown into air moving under gravity only is called a projectile.

Q2. What is the shape of projectile path?

Parabola.

Q3. What is horizontal acceleration in projectile motion?

Zero.

Q4. What happens to vertical velocity at highest point?

It becomes zero.

3. Short Answer Questions

Q1. Why is projectile motion called two-dimensional motion?

Projectile motion has both horizontal and vertical components of motion. Therefore it is called two-dimensional motion.

Q2. Why does horizontal velocity remain constant?

No horizontal force acts on the projectile. Therefore horizontal acceleration is zero and horizontal velocity remains constant.

Q3. State two assumptions in projectile motion.

1. Air resistance is neglected.
2. Only gravity acts on the projectile.

Q4. Write equations of horizontal and vertical positions.

x = (v₀ cosθ)t

y = (v₀ sinθ)t − ½gt²

4. Long Answer Questions

Q1. Explain projectile motion with equations.

Projectile motion is the motion of an object thrown into air under the influence of gravity only. It has two independent motions:

• Horizontal Motion: No acceleration acts horizontally. Therefore horizontal velocity remains constant.
• Vertical Motion: Gravity acts vertically downward. Therefore vertical velocity changes continuously.

Initial velocity components:

v₀x = v₀ cosθ
v₀y = v₀ sinθ

Position equations:

x = (v₀ cosθ)t
y = (v₀ sinθ)t − ½gt²

Velocity equations:

vx = v₀ cosθ
vy = v₀ sinθ − gt

The path followed by projectile is a parabola.

5. Assertion and Reason Questions

Q1. Assertion (A): At maximum height, vertical velocity becomes zero.

Reason (R): Gravity stops acting at maximum height.

Assertion is true but Reason is false.

Q2. Assertion (A): Horizontal velocity remains constant.

Reason (R): No horizontal acceleration acts on projectile.

Both Assertion and Reason are true and Reason correctly explains Assertion.

6. Fill in the Blanks

1. The path of projectile is __________.

Parabola

2. Horizontal acceleration in projectile motion is __________.

Zero

3. At highest point vertical velocity becomes __________.

Zero

4. Gravity acts in __________ direction.

Downward

7. Case Study Questions

A boy throws a ball with velocity v₀ at angle θ. The ball moves along a curved path and returns to ground.

Q1. What type of motion is this?

Projectile motion.

Q2. What is the shape of path?

Parabola.

Q3. Which force acts on the ball?

Gravitational force.

Q4. What happens to horizontal velocity?

It remains constant.

8. Statement Based Questions

1. Gravity acts horizontally in projectile motion.

False

2. Horizontal velocity remains constant.

True

3. Projectile motion is one-dimensional.

False

4. Vertical acceleration equals g.

True

9. Match the Columns

Column A	Column B
1. Horizontal acceleration	a. Parabola
2. Path of projectile	b. Zero
3. Vertical acceleration	c. g
4. Highest point	d. vy = 0

Answers:

1 → b
2 → a
3 → c
4 → d

10. HOTS Questions

Q1. Why does a projectile eventually fall to ground?

Gravity continuously pulls the projectile downward causing it to return to ground.

Q2. Can horizontal velocity become zero during projectile motion?

No. No horizontal acceleration acts on projectile.

11. Important Formulae

v₀x = v₀ cosθ

v₀y = v₀ sinθ

x = (v₀ cosθ)t

y = (v₀ sinθ)t − ½gt²

vx = v₀ cosθ

vy = v₀ sinθ − gt

Prepared for CBSE Class 11 Physics Students

Motion in a Plane with Constant Acceleration Notes for Beginners

 

MOTION IN A PLANE WITH CONSTANT ACCELERATION
│
├── 1. Motion in a Plane
│   │
│   ├── Two-dimensional motion
│   ├── Motion along x-axis and y-axis
│   └── Examples
│       ├── Ball thrown in air
│       ├── Flying bird
│       └── Airplane motion
│
├── 2. Constant Acceleration
│   │
│   ├── Acceleration remains same
│   ├── Magnitude constant
│   ├── Direction constant
│   └── Example
│       └── Gravity
│
├── 3. Velocity Equation
│   │
│   ├── Formula
│   │   └── v = v₀ + at
│   │
│   ├── Meaning
│   │   ├── Final velocity
│   │   ├── Initial velocity
│   │   └── Effect of acceleration
│   │
│   └── Velocity changes with time
│
├── 4. Velocity Components
│   │
│   ├── x-direction
│   │   └── vₓ = v₀ₓ + aₓt
│   │
│   └── y-direction
│       └── vᵧ = v₀ᵧ + aᵧt
│
├── 5. Position Equation
│   │
│   ├── Average velocity
│   │   └── (v₀ + v) / 2
│   │
│   ├── Position formula
│   │   └── r = r₀ + v₀t + ½at²
│   │
│   └── Depends on
│       ├── Initial position
│       ├── Initial velocity
│       ├── Acceleration
│       └── Time
│
├── 6. Position Components
│   │
│   ├── Along x-axis
│   │   └── x = x₀ + v₀ₓt + ½aₓt²
│   │
│   └── Along y-axis
│       └── y = y₀ + v₀ᵧt + ½aᵧt²
│
├── 7. Important Concept
│   │
│   ├── x-motion independent of y-motion
│   ├── Horizontal and vertical motions separate
│   └── Solve both directions independently
│
├── 8. Projectile Motion
│   │
│   ├── Horizontal motion
│   │   └── Constant velocity
│   │
│   └── Vertical motion
│       └── Acceleration due to gravity
│
└── 9. Key Points
    │
    ├── Two-dimensional motion
    ├── Constant acceleration equations
    ├── Separate x and y equations
    ├── Useful in projectile motion
    └── Easy to solve using components
Motion in a Plane with Constant Acceleration explained using simple formulas and projectile motion diagrams. 


- Dr.Sanjaykumar pawar

INTERNAL LINKS
Introduction to Vectors
Scalars and Vectors Notes
Projectile Motion Explained
Laws of Motion Notes
Kinematics Formula Sheet
Motion in a Straight Line
Velocity and Acceleration Basics
NCERT Class 11 Physics Notes
Important Physics Formulas
Two Dimensional Motion Examples

Motion in a Plane with Constant Acceleration

1. What is Motion in a Plane?

When an object moves in two directions at the same time (along x-axis and y-axis), it is called motion in a plane.

Examples:

• A ball thrown in air
• A flying bird
• An airplane moving in the sky

2. Constant Acceleration

Constant acceleration means acceleration does not change with time.

• Magnitude remains constant
• Direction remains constant

Example: Acceleration due to gravity near Earth.

3. Velocity Equation in Two Dimensions

Suppose:

• Initial velocity = v₀
• Final velocity = v
• Acceleration = a
• Time = t

From definition of acceleration:

a = (v - v₀) / t

Rearranging:

v = v₀ + at

Meaning:
Final velocity = Initial velocity + change due to acceleration

4. Velocity Components

Motion in a plane has two directions:

• x-direction (horizontal)
• y-direction (vertical)

Velocity Along x-axis

vₓ = v₀ₓ + aₓt

Velocity Along y-axis

vᵧ = v₀ᵧ + aᵧt

5. Position Equation in Two Dimensions

Suppose:

• Initial position = r₀
• Final position = r

Average velocity:

Average Velocity = (v₀ + v) / 2

Position equation becomes:

r = r₀ + v₀t + ½at²

Meaning:
Final position depends on initial position, velocity, acceleration and time.

6. Position Components

Position Along x-axis

x = x₀ + v₀ₓt + ½aₓt²

Position Along y-axis

y = y₀ + v₀ᵧt + ½aᵧt²

7. Important Concept

Motion in x-direction and y-direction are independent.

• Horizontal motion does not affect vertical motion.
• Vertical motion does not affect horizontal motion.
• Both motions can be solved separately.

8. Real Life Example: Projectile Motion

When a ball is thrown:

• Horizontal motion has constant velocity.
• Vertical motion has acceleration due to gravity.

9. Key Points to Remember

• Motion in a plane is two-dimensional motion.
• Acceleration remains constant.
• Velocity equation: v = v₀ + at
• Position equation: r = r₀ + v₀t + ½at²
• x and y motions are solved separately.

10. Short Summary

Two-dimensional motion can be divided into two one-dimensional motions. Separate equations are used for x-direction and y-direction. This concept is very useful in projectile motion.