Projectile Motion – Maximum Height & Range | Complete JEE Main Physics Guide

Projectile Motion – Maximum Height & Range: Complete JEE Main Guide

Projectile Motion JEE Main Physics ke important topics mein se ek hai. Is topic mein ek object ko initial velocity ke saath kisi angle par project kiya jaata hai aur uske baad object par mainly gravitational acceleration g act karta hai.

Projectile Motion ko solve karne ka sabse important concept hai velocity ko horizontal aur vertical components mein resolve karna. Once components clear ho jaate hain, Time of Flight, Maximum Height aur Horizontal Range ke formulas easily derive aur remember kiye ja sakte hain.

Is article mein hum Projectile Motion ke important concepts ko step-by-step samjhenge:

• Basic setup and velocity components
• Time of Flight
• Maximum Height
• Horizontal Range
• Maximum Range
• Maximum Height vs Maximum Range
• Complementary Angles
• Relation between H and R
• Special angles: 30°, 45° and 60°
• JEE Main solved question
• Common traps
• Quick revision formulas

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1. Projectile Motion – Basic Setup

Suppose ek projectile ko initial speed u se horizontal ke saath angle θ par project kiya gaya hai.

Projectile par downward acceleration:

Acceleration = g downward

Initial velocity ko do perpendicular components mein resolve karte hain.

Horizontal Component

u_x = u cosθ

Vertical Component

u_y = u sinθ

Yahi decomposition Projectile Motion ka foundation hai.

Horizontal direction mein acceleration zero hota hai, jabki vertical direction mein acceleration −g hota hai.

Direction	Initial Velocity	Acceleration
Horizontal	u cosθ	0
Vertical	u sinθ	−g

Isliye projectile motion ko essentially do independent motions ke combination ke roop mein treat kar sakte hain:

Horizontal → Uniform Motion

Vertical → Uniformly Accelerated Motion

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2. Time of Flight

Time of Flight projectile ke total airborne time ko represent karta hai, jab projectile same horizontal level par land karta hai jahan se launch hua tha.

Vertical velocity at time t:

v_y = u sinθ − gt

Landing ke time projectile ki vertical velocity:

v_y = −u sinθ

Therefore:

−u sinθ = u sinθ − gT

Rearranging:

gT = 2u sinθ

Hence:

T = 2u sinθ / g

Ye formula tab directly apply hota hai jab projectile same vertical level par land karta hai jahan se project hua tha.

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3. Maximum Height

Projectile jab maximum height par pahunchta hai, us moment vertical velocity zero ho jaati hai.

At maximum height:

v_y = 0

Vertical initial velocity:

u_y = u sinθ

Equation of motion:

v_y² = u_y² − 2gH

At maximum height:

0 = u²sin²θ − 2gH

Therefore:

2gH = u²sin²θ

Hence:

H = u²sin²θ / 2g

Important Observation

H ∝ sin²θ

For a fixed initial speed, angle increase hone par maximum height increase hoti hai.

Maximum height ke liye sin²θ ko maximum hona chahiye. Isliye θ = 90° par maximum height obtain hoti hai.

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4. Horizontal Range

Horizontal Range projectile ke starting point se landing point tak horizontal distance hota hai.

Horizontal velocity:

u_x = u cosθ

Horizontal direction mein acceleration zero hai, so horizontal velocity constant rehti hai.

Therefore:

Range = Horizontal Velocity × Time of Flight

So:

R = u cosθ × T

Time of flight:

T = 2u sinθ/g

Therefore:

R = u cosθ × 2u sinθ/g

R = 2u²sinθcosθ/g

Using:

2sinθcosθ = sin2θ

We get:

R = u²sin2θ/g

This is the most important formula for horizontal range.

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5. Maximum Range

For fixed initial velocity u:

R = u²sin2θ/g

Yahan u²/g constant hai. Range maximum tab hogi jab:

sin2θ = 1

Maximum value of sine = 1.

Therefore:

2θ = 90°

Hence:

θ = 45°

Therefore projectile maximum range tab cover karta hai jab projection angle:

45°

ho.

Maximum Range

At θ = 45°:

sin90° = 1

Therefore:

R_max = u²/g

Ye JEE Main ka extremely important result hai.

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6. Maximum Height vs Maximum Range

Maximum Height aur Maximum Range ko students often confuse kar dete hain. Dono different conditions par maximum hote hain.

Quantity	Formula	Maximum At
Maximum Height	H = u²sin²θ/2g	θ = 90°
Horizontal Range	R = u²sin2θ/g	θ = 45°

Key Concept:

Maximum Height ≠ Maximum Range

Height maximum:

θ = 90°

Range maximum:

θ = 45°

At 90°, projectile vertically upward jaata hai, isliye horizontal range zero hoti hai. At 45°, horizontal aur vertical components balanced manner mein range maximize karte hain.

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7. Complementary Angles and Same Range

Projectile Motion ka ek very useful JEE shortcut hai complementary angles.

Suppose:

θ₁ + θ₂ = 90°

Then:

θ₂ = 90° − θ₁

Range formula:

R = u²sin2θ/g

For complementary angles:

sin2θ₁ = sin(180° − 2θ₁)

Therefore:

R(θ) = R(90° − θ)

Hence complementary angles produce the same range, provided initial speed and launch/landing levels are the same.

Example

Consider:

30° and 60°

Since:

30° + 60° = 90°

Therefore:

R_30° = R_60°

But their maximum heights are different.

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8. Relation Between Maximum Height and Range

Projectile Motion mein H aur R ke beech ek important relation derive kiya ja sakta hai.

Maximum height:

H = u²sin²θ/2g

Range:

R = u²sin2θ/g

Using:

sin2θ = 2sinθcosθ

Therefore:

R = 2u²sinθcosθ/g

Now:

H/R = [u²sin²θ/2g] / [2u²sinθcosθ/g]

After simplification:

H/R = tanθ/4

Hence:

tanθ = 4H/R

Ye relation tab extremely useful hota hai jab question mein directly H aur R diye hon aur angle find karna ho.

JEE Shortcut

θ = tan⁻¹(4H/R)

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9. Special Case – θ = 30°

For:

θ = 30°

Range:

R = u²sin60°/g

Since:

sin60° = √3/2

Therefore:

R = √3u²/2g

Maximum height:

H = u²sin²30°/2g

Since:

sin30° = 1/2

Therefore:

H = u²/8g

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10. Special Case – θ = 45°

For:

θ = 45°

Range:

R = u²sin90°/g

Therefore:

R = u²/g

Ye maximum range hai.

Maximum height:

H = u²sin²45°/2g

Since:

sin45° = 1/√2

Therefore:

H = u²/2g × 1/2

H = u²/4g

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11. Special Case – θ = 60°

For:

θ = 60°

Range:

R = u²sin120°/g

Since:

sin120° = sin60° = √3/2

Therefore:

R = √3u²/2g

Maximum height:

H = u²sin²60°/2g

Since:

sin60° = √3/2

Therefore:

H = 3u²/8g

Important Observation

R_30° = R_60°

But:

H_60° > H_30°

Higher angle projectile zyada height tak jaata hai, while complementary angles same range cover kar sakte hain.

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12. JEE Main Solved Question

Question

A projectile is projected with speed 20 m/s at an angle 30° with the horizontal. Take g = 10 m/s². Find:

(i) Maximum Height

(ii) Horizontal Range

Solution – Maximum Height

Formula:

H = u²sin²θ/2g

Given:

u = 20 m/s

θ = 30°

g = 10 m/s²

Therefore:

H = 20² × (1/2)² / (2 × 10)

H = 400 × 1/4 / 20

H = 100/20

H = 5 m

Therefore:

Maximum Height = 5 m

Solution – Horizontal Range

Formula:

R = u²sin2θ/g

Therefore:

R = 400 × sin60° / 10

Since:

sin60° = √3/2

Therefore:

R = 40 × √3/2

R = 20√3 m

Hence:

Maximum Height = 5 m

Range = 20√3 m

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13. Important Concept – Horizontal Velocity

Projectile Motion mein horizontal acceleration zero hota hai, assuming air resistance ignore kiya gaya hai.

Therefore horizontal velocity remains constant:

u_x = u cosθ = constant

Iska matlab ye nahi hai ki projectile ki total velocity constant hai. Vertical velocity continuously change hoti hai because gravity act kar rahi hai.

Vertical velocity:

v_y = u sinθ − gt

At the highest point:

v_y = 0

But horizontal velocity still remains:

v_x = u cosθ

So highest point par total velocity zero nahi hoti, unless horizontal component bhi zero ho.

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14. Common JEE Traps

Trap 1 – Maximum Range at 90°

Wrong:

Maximum range → 90°

Correct:

Maximum range → 45°

Trap 2 – Maximum Height at 45°

45° maximum range ke liye hota hai, maximum height ke liye nahi.

Maximum Height → 90°

Trap 3 – Complementary Angles Have Same Height

Wrong.

Complementary angles same range provide karte hain, but maximum heights generally different hoti hain.

θ₁ + θ₂ = 90° → Same Range

Trap 4 – Horizontal Velocity Changes

Air resistance ignore karne par:

u_x = u cosθ = constant

Horizontal velocity change nahi hoti.

Trap 5 – Highest Point Par Vertical Velocity Maximum

Exactly opposite.

Highest point par:

v_y = 0

Vertical velocity zero hoti hai.

Trap 6 – Range Formula Without Checking Conditions

R = u²sin2θ/g standard same-level launch and landing case ke liye hai. Agar projectile different height se launch ya land kare, to direct formula apply nahi karna chahiye.

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15. Important Formula Sheet

Quantity	Formula
Horizontal velocity	ux = u cosθ
Vertical velocity	uy = u sinθ
Time of Flight	T = 2u sinθ/g
Maximum Height	H = u²sin²θ/2g
Horizontal Range	R = u²sin2θ/g
Maximum Range	Rmax = u²/g
Angle for Maximum Range	θ = 45°
H-R Relation	H/R = tanθ/4
Angle from H and R	tanθ = 4H/R

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16. JEE Main Practice Questions

Question 1

A projectile is projected at 45° with speed u. Find its maximum range.

Answer: u²/g

Question 2

Two projectiles are projected with the same speed at 30° and 60°. Compare their horizontal ranges.

Answer: Same range.

Question 3

For a projectile, maximum height is H and range is R. Find the angle of projection.

Answer: θ = tan⁻¹(4H/R)

Question 4

At what angle is the horizontal range maximum for a fixed initial speed?

Answer: 45°

Question 5

What is the vertical velocity of a projectile at its highest point?

Answer: Zero.

Question 6

If the launch angle is changed from 30° to 60° while speed remains the same, what happens to the range?

Answer: Range remains the same.

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17. How to Solve Projectile Motion Questions Quickly

JEE Main mein Projectile Motion questions solve karne ke liye sabse pehle identify karo ki question exactly kya pooch raha hai.

1. Time of Flight: Use T = 2u sinθ/g.
2. Maximum Height: Use H = u²sin²θ/2g.
3. Range: Use R = u²sin2θ/g.
4. Maximum Range: Immediately θ = 45° and R_max = u²/g.
5. Complementary angles: θ and 90°−θ → same range.
6. H and R given: Use tanθ = 4H/R.

Question mein agar highest point mention ho, immediately remember:

v_y = 0

Agar same level landing mention ho, Time of Flight aur standard Range formulas directly use kiye ja sakte hain.

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18. One-Minute Revision

Projectile Motion revise karte waqt ye formulas sabse pehle recall karo:

u_x = u cosθ

u_y = u sinθ

T = 2u sinθ/g

H = u²sin²θ/2g

R = u²sin2θ/g

R_max = u²/g

θ for R_max = 45°

Complementary angles:

θ₁ + θ₂ = 90° → Same Range

H-R relation:

tanθ = 4H/R

Highest point:

v_y = 0

Horizontal direction:

u_x = constant

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19. Final Revision Box 🔥

Initial Horizontal Component:

u_x = u cosθ

Initial Vertical Component:

u_y = u sinθ

Time of Flight:

T = 2u sinθ/g

Maximum Height:

H = u²sin²θ/2g

Horizontal Range:

R = u²sin2θ/g

Maximum Range:

R_max = u²/g

Angle for Maximum Range:

θ = 45°

Complementary Angles:

θ₁ + θ₂ = 90° → Same Range

H-R Relation:

tanθ = 4H/R

Highest Point:

v_y = 0

Super Trick:

Height → sin²θ

Range → sin2θ

Maximum Range → 45°

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20. Download / View Projectile Motion PDF

Detailed Projectile Motion notes aur revision material ke liye PDF yahan embed kiya ja sakta hai:

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21. Frequently Asked Questions (FAQs)

Q1. Projectile Motion mein horizontal velocity kya hoti hai?

Initial horizontal velocity:

u cosθ

Air resistance ignore karne par ye constant rehti hai.

Q2. Projectile ki maximum height ka formula kya hai?

H = u²sin²θ/2g

Q3. Horizontal range ka formula kya hai?

Same launch aur landing level ke case mein:

R = u²sin2θ/g

Q4. Maximum range kis angle par hoti hai?

Fixed initial speed ke liye maximum range 45° par hoti hai.

Q5. Kya 30° aur 60° par range same hoti hai?

Haan, same initial speed aur same launch/landing level ke case mein 30° aur 60° complementary angles hain, isliye range same hoti hai.

Q6. Highest point par vertical velocity kya hoti hai?

Zero.

Q7. Kya highest point par projectile ki total velocity zero hoti hai?

Generally nahi. Highest point par sirf vertical component zero hota hai. Horizontal component u cosθ remain karta hai.

Q8. H aur R diye hone par angle kaise find karenge?

Use:

tanθ = 4H/R

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Final Thoughts

Projectile Motion ko difficult samajhne ki zarurat nahi hai. Is topic ka core idea sirf velocity ko horizontal aur vertical components mein resolve karna hai.

Ek baar ye clear ho gaya, to important formulas automatically connected ho jaate hain:

u cosθ → Horizontal Motion

u sinθ → Vertical Motion

Phir:

T = 2u sinθ/g

H = u²sin²θ/2g

R = u²sin2θ/g

JEE Main ke liye sabse important concepts ko ek saath yaad rakho:

Maximum Height → 90°

Maximum Range → 45°

Complementary Angles → Same Range

Highest Point → Vertical Velocity = 0

H and R → tanθ = 4H/R

In formulas aur concepts ko numerical practice ke saath revise karte raho. Projectile Motion ke direct questions phir kaafi quickly solve kiye ja sakte hain.

Concept clear rakho, components identify karo, correct formula choose karo, aur JEE Main mein traps se bacho.