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