Concept 03 · Motion in a Plane

Projectile Motion progress0 of 1 completed

Projectile Motion

Follow an object through the air and discover why its path is curved.

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3Learning ideas 1Questions 1Videos

Projectile trajectory showing maximum height and horizontal range

Understand

Build your foundation

See the idea first. Then give the mathematics a meaning.

01 · Start with why

Why this concept matters

Learn how horizontal and vertical components combine to produce projectile motion.

02 · Read the visual

See the relationship

Understand and solve projectile motion problems using component-wise analysis.

Trajectory

Explore the visual idea, follow an example, then try it yourself.

Range

Explore the visual idea, follow an example, then try it yourself.

Time of flight

Explore the visual idea, follow an example, then try it yourself.

?Understanding checkpointCan you explain the idea without using a formula?

0 basic questions

Apply

Practise and solve

Use the concept across the complete NCERT learning path.

NCERT practice

Turn clarity into confidence

Work through published questions connected to this concept.

1questions

Start Practice →

Worked thinking

Connect idea to method

Move from the physical picture to a worked method, one decision at a time.

NUMERICAL3.13Apply the concept →

Master

Think beyond the textbook

Build the depth needed for unfamiliar JEE and NEET applications.

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NEET

Fast conceptual decisions and accurate application.

Exam practice

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JEE Main

Multi-step reasoning with speed and precision.

Deeper problems

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JEE Advanced

Challenge assumptions and connect multiple ideas.

Selected challenges

Exam questions will appear here as they are reviewed and published.

Key formulas

Meaning before memory

Every formula belongs to a visual idea.

Horizontal Component of Velocityu_x=u\cos\theta

Horizontal launch-velocity component when the projection angle is measured from the positive x-axis.

Vertical Component of Velocityu_y=u\sin\theta

Vertical launch-velocity component when the projection angle is measured from the positive x-axis.

Time of FlightT=\frac{2u\sin\theta}{g}

Total flight time when the projectile lands at the same height from which it was launched.

Maximum HeightH=\frac{u^2\sin^2\theta}{2g}

Highest vertical displacement above the launch point.

Horizontal RangeR=\frac{u^2\sin 2\theta}{g}

Horizontal distance travelled when launch and landing heights are the same.

Projectile Trajectoryy=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}

Equation of the parabolic path when air resistance is ignored.

Video lessons

Watch the idea unfold

Short lessons placed beside the ideas they explain.

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Lesson 01

NCERT Class 11 Physics Q3.12 | Why 45° Doesn't Work? | Projectile Motion | Chapter 3 | CBSE

Quick revision

Make the idea stick

Recall the essentials without rereading the whole lesson.

ƒFormula recapRelationships in one place↻Visual recapRevisit the key ideas▶Watch the series1 published lesson

Concept journey

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