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1. Re-visiting Motion in a Plane 🎯

In one-dimensional motion, everything happens along a single straight line. In two-dimensional motion (motion in a plane), an object’s position changes in both x and y directions simultaneously.

Why this chapter is a JEE favourite?

• Combines kinematics + vectors (both are core JEE topics)
• Great for testing concepts + calculation speed
• Often linked with questions on:
  • Projectile motion
  • Relative velocity in 2D
  • River-boat and airplane-wind problems

In 2D motion, we usually:

• Choose coordinate axes (x-horizontal, y-vertical)
• Break the motion into two independent parts:
  • Horizontal motion
  • Vertical motion
• Use 1D kinematics equations separately along each axis

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2. Vector Breakdown: The Real Game-changer 🧭

Any vector in a plane (displacement, velocity, acceleration) can be resolved into two perpendicular components.

If a vector has magnitude $v$ and makes an angle $\theta$ with the positive x-axis, then:

$$v_x = v \cos \theta$$ $$v_y = v \sin \theta$$

We usually write:

$$\vec{v} = v_x \hat{i} + v_y \hat{j}$$

Quick Concept Table 📋

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3. Projectiles in a Plane: The Heart of Set-2 ⚽

Projectile motion is 2D motion under gravity when:

• Gravity is the only acceleration
• Air resistance is neglected
• Acceleration is constant, directed vertically downward

Typical example: a ball thrown at an angle to the horizontal.

Assume:

• Initial speed = $u$
• Angle of projection with horizontal = $\theta$
• Acceleration due to gravity = $g$ downward
• Origin at point of projection
• Upward direction = positive y

Velocity Components

$$u_x = u \cos \theta$$ $$u_y = u \sin \theta$$

Horizontal motion:

• No horizontal acceleration (assuming no air resistance)
• So $a_x = 0$

Vertical motion:

• Constant acceleration downward
• So $a_y = -g$

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4. Equation of Trajectory 🏹

Horizontal motion:

$$x = u_x \, t = u \cos \theta \cdot t$$

Vertical motion:

$$y = u_y \, t - \frac{1}{2} g t^2$$

Substitute $t$ from first equation:

From $x = u \cos \theta \, t$,

$$t = \frac{x}{u \cos \theta}$$

Now put this in expression for $y$:

$$y = u \sin \theta \left(\frac{x}{u \cos \theta}\right) - \frac{1}{2} g \left(\frac{x}{u \cos \theta}\right)^2$$

Simplify:

$$y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$$

This is of the form:

$$y = a x + b x^2$$

which represents a parabola. So projectile path is parabolic.

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5. Key Results You Must Memorise (and Understand) 📌

Consider a projectile launched from the ground with speed $u$ at angle $\theta$.

5.1 Time of Flight

When it lands back on the same level:

$$T = \frac{2 u \sin \theta}{g}$$

5.2 Maximum Height

$$H = \frac{u^2 \sin^2 \theta}{2 g}$$

5.3 Horizontal Range

$$R = \frac{u^2 \sin 2\theta}{g}$$

5.4 Angle for Maximum Range

Range is maximum when $\sin 2\theta = 1$.

So:

$$2\theta = 90^\circ \Rightarrow \theta = 45^\circ$$

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Quick Revision Box ✏️

• Horizontal velocity is constant (if no air resistance).
• Vertical velocity changes due to gravity.
• At the highest point:
  • Vertical component of velocity = 0
  • Horizontal component still = $u \cos \theta$
• Time to reach maximum height:

$$t_\text{up} = \frac{u \sin \theta}{g}$$

• Time of flight = $2 t_\text{up}$ for symmetric projectile (landing at same height).

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6. JEE-style Solved Example (Projectile) 🧮

Example 1 (Oblique Projection from Ground)
A projectile is fired with speed 20 m/s at an angle of 30° with the horizontal. Take $g = 10$ m/s². Find:

1. Time of flight
2. Maximum height
3. Horizontal range

Step 1: Write given data

• $u = 20$ m/s
• $\theta = 30^\circ$
• $g = 10$ m/s²

Use:

• $\sin 30^\circ = 1/2$
• $\cos 30^\circ = \sqrt{3} / 2$
• $\sin 60^\circ = \sqrt{3} / 2$

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6.1 Time of Flight

Formula:

$$T = \frac{2 u \sin \theta}{g}$$

Substitute:

$$T = \frac{2 \times 20 \times 1/2}{10} = \frac{20}{10} = 2 \text{ s}$$

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6.2 Maximum Height

Formula:

$$H = \frac{u^2 \sin^2 \theta}{2 g}$$

Compute:

$$H = \frac{20^2 \times (1/2)^2}{2 \times 10}$$ $$H = \frac{400 \times 1/4}{20} = \frac{100}{20} = 5 \text{ m}$$

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6.3 Horizontal Range

Formula:

$$R = \frac{u^2 \sin 2\theta}{g}$$

Here $2\theta = 60^\circ$.

So:

$$R = \frac{20^2 \times \sqrt{3}/2}{10}$$ $$R = \frac{400 \times \sqrt{3}/2}{10} = \frac{200 \sqrt{3}}{10} = 20 \sqrt{3} \text{ m}$$

Final answers:

• Time of flight = 2 s
• Maximum height = 5 m
• Range = $20 \sqrt{3}$ m

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7. Relative Motion in a Plane: Boats & Planes 🚤✈️

Relative velocity is heavily tested in JEE, especially in 2D.

If an object A moves with velocity $\vec{v}_A$ and object B moves with velocity $\vec{v}_B$, then velocity of A relative to B is:

$$\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$$

7.1 River-Boat Problems

• River flows with velocity of water $\vec{v}_w$ (usually along x-axis).
• Boat’s velocity relative to water is $\vec{v}_{b/w}$.
• Actual velocity of boat relative to ground:

$$\vec{v}_{b/g} = \vec{v}_{b/w} + \vec{v}_w$$

Typical cases:

1. Boat aimed straight across the river
2. Boat aimed at some angle to land at a point directly opposite

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Visualising the Motion (Diagram Description) 🖼️

Imagine a top view:

• River banks are horizontal lines.
• Water flows to the right with velocity arrows along the x-direction.
• Boat’s intended direction is an arrow making angle with vertical.
• The resultant velocity is the diagonal of a parallelogram formed by velocity vectors of boat and water.

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7.2 Solved Example (Boat Problem)

Example 2
A river flows east with speed 3 m/s. A boat can move at 5 m/s in still water. The boat is steered so that its velocity relative to water is due north.

1. What is the boat’s actual velocity relative to ground (magnitude and direction)?
2. How far downstream will it drift while crossing a 100 m wide river?

Step 1: Understand the vectors

Let:

• East = +x direction
• North = +y direction

Given:

• Velocity of water relative to ground:

$$\vec{v}_w = 3 \hat{i} \text{ m/s}$$

• Velocity of boat relative to water:

$$\vec{v}_{b/w} = 5 \hat{j} \text{ m/s}$$

So:

$$\vec{v}_{b/g} = \vec{v}_{b/w} + \vec{v}_w = 3 \hat{i} + 5 \hat{j}$$

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7.2.1 Magnitude of Resultant Velocity

Magnitude:

$$v_{b/g} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \text{ m/s}$$

Approx:

$$v_{b/g} \approx 5.83 \text{ m/s}$$

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7.2.2 Direction of Motion

Let angle $\phi$ be measured from the north direction towards east.

Then:

$$\tan \phi = \frac{\text{east component}}{\text{north component}} = \frac{3}{5}$$

So:

$$\phi = \tan^{-1} \left(\frac{3}{5}\right)$$

This is roughly 31° east of north.

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7.2.3 Time to Cross the River

Width of river = 100 m, which is along north-south (y-direction).

Component of velocity perpendicular to banks (along y):

• It is 5 m/s (the boat’s entire relative-to-water speed is north).

Time to cross:

$$t = \frac{\text{distance across}}{\text{perpendicular speed}} = \frac{100}{5} = 20 \text{ s}$$

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7.2.4 Drift Downstream

Downstream drift is due to the x-component of boat’s ground velocity.

This is 3 m/s (same as river speed).

Distance drifted:

$$x = v_x \, t = 3 \times 20 = 60 \text{ m}$$

Answers:

• Boat’s speed = $\sqrt{34}$ m/s ≈ 5.83 m/s, direction ≈ 31° east of north
• Downstream drift = 60 m

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8. Common Mistakes in 2D Motion (JEE Trap Alert) 🚨

1. Forgetting that motions along x and y are independent

   • Students often try to mix accelerations or use wrong signs.

2. Using wrong trigonometric functions

   • Confusing $\sin$ and $\cos$ when resolving velocity components.

3. Taking gravity as positive upward

   • In most questions, choose upward as positive, then $a_y = -g$.

4. Not distinguishing between angle with horizontal and vertical

   • Always note whether angle is given with the horizontal or vertical.

5. Assuming range formula always works

   • Formula $R = u^2 \sin 2\theta / g$ works only when:
     • Launch and landing heights are same
     • Constant g, negligible air resistance

6. Relative motion sign errors

   • Velocity of A relative to B is $\vec{v}_A - \vec{v}_B$, not the other way around.

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9. Mixed Concept Example (Projectile + Relative Motion) 🔗

Example 3 (Airplane with Wind)
An airplane has an airspeed (speed relative to air) of 200 m/s. There is a wind blowing from west to east at 50 m/s. The pilot wants to move exactly towards the north.

1. At what angle should the airplane head relative to north?
2. What will be its ground speed?

Step 1: Set axes

• East = +x
• North = +y

Wind velocity:

$$\vec{v}_w = 50 \hat{i} \text{ m/s}$$

Let the airplane’s velocity relative to air have magnitude 200 m/s and be at angle $\alpha$ west of north (towards negative x).

So components of plane relative to air:

• x-component: $-200 \sin \alpha$
• y-component: $200 \cos \alpha$

So:

$$\vec{v}_{p/a} = -200 \sin \alpha \, \hat{i} + 200 \cos \alpha \, \hat{j}$$

Ground velocity:

$$\vec{v}_{p/g} = \vec{v}_{p/a} + \vec{v}_w$$ $$\vec{v}_{p/g} = (-200 \sin \alpha + 50) \hat{i} + 200 \cos \alpha \hat{j}$$

For motion to be exactly north, x-component must be zero:

$$-200 \sin \alpha + 50 = 0$$

So:

$$200 \sin \alpha = 50 \Rightarrow \sin \alpha = \frac{1}{4}$$

Thus:

$$\alpha = \sin^{-1} \left(\frac{1}{4}\right)$$

Airplane should head $\alpha$ west of north.

Now y-component (ground speed, since no x component):

$$v_{p/g} = 200 \cos \alpha$$

But:

$$\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - 1/16} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$$

So:

$$v_{p/g} = 200 \times \frac{\sqrt{15}}{4} = 50 \sqrt{15} \text{ m/s}$$

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10. Real-life Connections 🌍

• Sports: Basketball arcs, football free kicks, long jump trajectories are all projectiles. Coaches use these ideas to optimize angles and speeds.
• Military & Space: Shell trajectories, missile guidance, even satellite launches in their early phases depend on 2D and 3D motion calculations.
• Aviation & Shipping: Pilots and ship captains constantly adjust for wind and water currents using relative velocity concepts.
• Rescue Operations: Dropping supplies from helicopters, firefighting aircrafts – they must calculate releasing point using projectile motion.

Visualising these situations strengthens your conceptual clarity and helps you recall formulas during exams.

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11. Exam Strategy for JEE Kinematics in a Plane 🧠

• Step 1: Draw a clear diagram
  Include axes, components of velocity, angles, and directions of acceleration.

• Step 2: Resolve into components
  Write separate equations for x and y directions.

• Step 3: Decide which equations to use
  Typical 1D kinematics equations:

  • $v = u + a t$
  • $s = u t + \frac{1}{2} a t^2$
  • $v^2 = u^2 + 2 a s$

• Step 4: Use symmetry whenever possible
  In projectile problems returning to same height:

  • Time to go up = time to come down
  • Path is symmetric about highest point

• Step 5: Check units and angle
  Convert degrees to radians only if required (usually in theory or graphs), otherwise keep in degrees for numeric questions.

• Step 6: Don’t blindly use formulas
  First verify if the conditions of the standard formula are satisfied (e.g., same level for range).

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12. Ultra-Fast Revision Table ⚡

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13. Ready to Test Yourself? ✅

Put your understanding of vectors, projectiles, and relative motion to the test with carefully curated questions based on the same concepts discussed above.