Kinotics

Relative Motion Between Two Projectiles

Relative motion is one of the most important concepts in classical mechanics for JEE and NEET projectile problems.
Instead of studying two projectiles separately, we observe one projectile from the frame of the other.

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Basic Idea of Relative Motion

If two projectiles have velocities:

v₁ = velocity of projectile 1
v₂ = velocity of projectile 2

Then relative velocity of projectile 1 with respect to projectile 2 is:

v₁₂ = v₁ − v₂

Similarly, relative position:

r₁₂ = r₁ − r₂

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Most Important Concept

Both projectiles experience the same gravitational acceleration:

a₁ = a₂ = −g ĵ

Therefore relative acceleration becomes:

a₁₂ = a₁ − a₂ = 0

This means:

• Relative acceleration is zero
• Relative velocity remains constant
• Relative motion becomes uniform straight-line motion

This is the key shortcut for JEE and NEET.

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Relative Position Equation

If initial relative position is:

r₀ = r₁(0) − r₂(0)

Then relative position after time t becomes:

r₁₂(t) = r₀ + v₁₂ t

Notice:

There is no t² term because relative acceleration is zero.

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Projectile Equations

Projectile 1

Initial speed = u₁
Projection angle = θ₁

Horizontal position:

x₁ = u₁ cosθ₁ · t

Vertical position:

y₁ = u₁ sinθ₁ · t − (1/2)gt²

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Projectile 2

Initial speed = u₂
Projection angle = θ₂

Horizontal position:

x₂ = u₂ cosθ₂ · t

Vertical position:

y₂ = u₂ sinθ₂ · t − (1/2)gt²

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Relative Coordinates

Relative horizontal coordinate:

x₁₂ = x₁ − x₂

x₁₂ = (u₁ cosθ₁ − u₂ cosθ₂)t

Relative vertical coordinate:

y₁₂ = y₁ − y₂

y₁₂ = (u₁ sinθ₁ − u₂ sinθ₂)t

Notice carefully:

The gravitational terms cancel completely.

Therefore relative motion becomes linear.

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Condition for Collision

Two projectiles collide when their relative position becomes zero.

Condition:

r₁₂ = 0

or

r₀ + v₁₂ t = 0

This means:

They must come on the same straight line

Relative velocity must point toward the other projectile

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Physical Interpretation

Although both projectiles move along parabolic paths relative to the ground:

From the frame of one projectile, the other appears to move in a straight line with constant velocity.

This happens because gravity affects both projectiles equally.

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Important Results for JEE & NEET

Relative acceleration:

a₁₂ = 0

Relative velocity:

v₁₂ = v₁ − v₂

Relative position:

r₁₂ = r₀ + v₁₂ t

Collision condition:

r₁₂ = 0

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Quick Revision

• Gravity cancels in relative motion
• Relative acceleration becomes zero
• Relative motion is straight-line motion
• Relative velocity remains constant
• Very important for collision and meeting problems in projectile motion

“PROJECTILE MOTION ALONG AN INCLINED PLANE”

Below the title, a large central physics diagram shows a projectile launched from the bottom of a sloped surface. The inclined plane makes angle β with the horizontal ground. The projectile is projected with initial speed u at angle α relative to the incline. A dashed blue parabolic trajectory lands back on the slope at point P.

Clearly labeled vectors and symbols: u = initial velocity
α = angle of projection with incline
β = inclination angle
R = range along incline
T = time of flight

Coordinate axes x and y are visible.

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SECTION 1 : RESOLVING MOTION

Green-colored educational box explaining rotated coordinate axes.

Axes are chosen: • parallel to incline • perpendicular to incline

Velocity components shown clearly:

u_parallel = u cos α

u_perpendicular = u sin α

Gravity components shown with arrows:

g_parallel = g sin β

g_perpendicular = g cos β

All vectors are color coded with clean directional arrows.

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SECTION 2 : EQUATIONS OF MOTION

Purple concept panel containing equations in rotated coordinates.

Perpendicular motion equation:

y' = u sin α · t − (1/2) g cos β · t²

Parallel motion equation:

x' = u cos α · t − (1/2) g sin β · t²

Equations are centered and fully visible without clipping.

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SECTION 3 : TIME OF FLIGHT

Orange derivation box showing projectile returns to incline when:

y' = 0

Derivation steps are displayed neatly.

Final highlighted equation:

T = (2u sin α) / (g cos β)

Large glowing formula box with clean typography.

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SECTION 4 : RANGE ALONG INCLINED PLANE

Blue-colored formula section explaining distance measured along the slope.

Final range equation displayed clearly:

R = [2u² sin α cos(α + β)] / [g cos² β]

Equation is fully visible on one line.

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SECTION 5 : MAXIMUM RANGE CONDITION

Red-highlighted box with target icon.

Condition for maximum range:

α = 45° − (β / 2)

Special notes: • If β = 0°, then α = 45° • As incline angle increases, optimum launch angle decreases

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SECTION 6 : IMPORTANT NOTES

Yellow information box listing key concepts:

• Time of flight depends on g cos β
• Range is measured along the incline
• Larger incline angle decreases range
• Larger incline angle decreases flight time
• Rotated axes simplify projectile motion problems

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SECTION 7 : REAL-LIFE APPLICATIONS

Small colorful illustrations showing: • football kicked uphill • missile launched from mountains • ski jump trajectory • ball thrown on a slope

Aircraft Motion in a Plane is an important topic of Motion in Two Dimensions for JEE Main and NEET. These problems are based on relative velocity and vector addition. The actual motion of the aircraft depends on both the velocity of the aircraft and the velocity of the wind.

Basic relative velocity equation:

V(PG) = V(PA) + V(AG)

Where:

V(PG) = velocity of plane relative to ground

V(PA) = velocity of plane relative to air

V(AG) = velocity of air or wind relative to ground

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CASE 1 : WIND ALONG THE DIRECTION OF PLANE

Same direction:

VR = VP + VW

Opposite direction:

VR = VP - VW

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CASE 2 : WIND PERPENDICULAR TO AIRCRAFT MOTION

Resultant velocity:

VR = √(VP² + VW²)

Direction:

tan(theta) = VW / VP

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CASE 3 : SHORTEST PATH CONDITION

To reach exactly at the destination point, the pilot tilts the aircraft against the wind.

Condition:

VP sin(theta) = VW

Effective velocity:

Veffective = √(VP² - VW²)

Time to travel distance d:

t = d / √(VP² - VW²)

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FASTEST TIME CONDITION

If the pilot does not compensate for the wind:

Aircraft drifts with wind

Time becomes minimum

Path is not straight

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NUMERICAL EXAMPLE 1

A plane flies north at 200 km/h. Wind blows east at 150 km/h.

Resultant velocity:

VR = √(200² + 150²)

VR = 250 km/h

Direction:

tan(theta) = 150 / 200

theta ≈ 37°

Plane moves 37° east of north.

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NUMERICAL EXAMPLE 2

Aircraft speed in still air:

VP = 300 km/h

Wind speed:

VW = 100 km/h

Required heading angle:

300 sin(theta) = 100

sin(theta) = 1/3

theta ≈ 19.5°

Pilot should tilt the aircraft 19.5° against wind.

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IMPORTANT JEE MAIN & NEET TRICKS

Aircraft problems are exactly similar to river-boat problems.

Always draw vector diagrams first.

Use vector addition carefully.

For shortest path → tilt against wind.

For minimum time → do not fight the wind.

Most questions use:

Relative velocity

Pythagoras theorem

Trigonometry

Vector triangle method

🌧️ RAIN MAN PROBLEM — RELATIVE VELOCITY EXPLAINED 🌧️

One of the most important applications of Relative Velocity in JEE Main and NEET physics is the famous Rain Man Problem. This concept explains why rain appears tilted to a moving observer even when it is actually falling vertically.

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🔹 THE SITUATION

A man runs horizontally with speed vm while rain falls vertically downward with speed vr.

To the man, the rain does not appear vertical.
Instead, it appears to come at an angle θ.

This happens because the observer is moving, creating a relative velocity between the rain and the man.

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🔹 RELATIVE VELOCITY CONCEPT

The apparent velocity of rain with respect to the moving man is:

Vrm = Vr − Vm

Where:

• Vr = velocity of rain
• Vm = velocity of man
• Vrm = apparent velocity of rain relative to man

This relative velocity determines the direction in which the rain appears to fall.

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🔹 VECTOR DIAGRAM UNDERSTANDING

In the diagram:

• Rain velocity acts vertically downward
• Man’s velocity acts horizontally
• Apparent rain direction is the diagonal resultant vector

This forms a right triangle:

• Horizontal side = vm
• Vertical side = vr
• Resultant side = Vrm

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🔹 IMPORTANT FORMULA

Since the angle is measured with the vertical:

tanθ = Horizontal Component / Vertical Component

Therefore:

tanθ = vm / vr

This is the key formula used in most rain-man problems.

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🔹 GIVEN CONDITION

Given:

tanθ = 9 / 16

Comparing with:

tanθ = vm / vr

We get:

vm / vr = 9 / 16

Hence:

vm : vr = 9 : 16

✅ Final Answer:

Ratio of speed of man to speed of rain = 9 : 16

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🔹 PHYSICAL INTERPRETATION

• 9 represents the horizontal motion of the man
• 16 represents the vertical speed of rain

Meaning:

For every 9 units of man’s speed, rain falls 16 units vertically.

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🔹 WHY RAIN APPEARS TILTED

If the man were standing still, rain would appear vertical.

But when the man moves forward:

• His frame of reference changes
• Rain seems to come from the opposite direction

So if the man runs rightward, rain appears tilted toward him from the front.

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🔹 EXAM TRICK 🚀

Remember:

“Rain always appears opposite to the observer’s motion.”

This shortcut helps solve direction questions quickly in JEE and NEET.

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🔹 SPECIAL CASES

✔ If vm = 0

θ = 0°

Rain appears vertical.

✔ If man runs faster

vm increases ⇒ θ increases

Rain appears more tilted.

✔ If wind gives rain a horizontal component

Use relative horizontal velocity before applying the tangent formula.

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🔹 CONNECTION WITH OTHER TOPICS

This problem is directly related to:

• Relative Velocity
• Vector Subtraction
• Motion in a Plane
• Frames of Reference
• Projectile Motion Concepts

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📘 JEE & NEET KEY TAKEAWAY

The Rain Man Problem is fundamentally a relative motion problem where the apparent direction of rain depends on the motion of the observer.

Mastering this concept makes many velocity and vector problems much easier.