Learn at My Place/Competitive Exams/NEET/Physics/Motion in a Plane

NEET Physics — Chapter 4

Motion in a Plane

Master 2D kinematics for NEET — projectile motion, uniform and non-uniform circular motion, relative velocity in 2D, and angular quantities with their kinematic equations.

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NEET Intro

Why Motion in a Plane Is Critical for NEET

Motion in a plane — particularly projectile motion and circular motion — is one of the highest-yield chapters in NEET Physics. Questions from this chapter appear every single year, often as standalone MCQs but also embedded in mechanics questions involving inclined planes, strings, and rotating systems.

Projectile motion is elegant because the independence of x and y motions reduces a 2D problem to two simultaneous 1D problems. Circular motion introduces the concept of centripetal acceleration, which recurs in gravitation, electromagnetism (charged particle in a field), and fluid mechanics.

Students who master the three key formulas (T, H, R), the trajectory equation, and the five common exam traps will handle virtually every NEET question from this chapter with confidence.

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Section A

Notes: Motion in a Plane

Original teaching copy for Learn at My Place

2D motion fundamentals Projectile motion: launch from ground Key projectile formulas (T, H, R)Equation of trajectory Projectile special cases + SVG Uniform circular motion Non-uniform circular motion Relative velocity in 2D Angular quantities NEET exam traps

1. 2D Motion Fundamentals

When motion occurs in a plane, we extend the 1D kinematic framework to two perpendicular directions simultaneously. The position, velocity, and acceleration each become two-component vectors.

\vec{r} = x\hat{i} + y\hat{j}

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

\vec{a} = a_x\hat{i} + a_y\hat{j}

The crucial insight is the principle of independence of motions: the x-component of motion is entirely independent of the y-component. This means the kinematic equations $v = u + at$ , $s = ut + \frac{1}{2}at^2$ , and $v^2 = u^2 + 2as$ are applied separately in each direction.

NEET tip: In 2D problems, always split initial velocity into x and y components first. Then apply 1D equations independently along each axis. Time is the linking variable — it is the same in both directions at any instant.

The magnitude of the resultant velocity at any instant is $v = \sqrt{v_x^2 + v_y^2}$ , and its direction is $\theta = \tan^{-1}(v_y/v_x)$ with the x-axis.

2. Projectile Motion: Launch from Ground

A projectile is any object launched with an initial velocity and then moving under gravity alone (no air resistance). If launched at speed $u$ at angle $\theta$ above the horizontal, the initial velocity components are:

u_x = u\cos\theta, \quad u_y = u\sin\theta

Horizontal motion (no acceleration, $a_x = 0$ ):

v_x = u\cos\theta \ (\text{constant throughout flight})

x = u\cos\theta \cdot t

Vertical motion (acceleration $g$ downward, $a_y = -g$ ):

v_y = u\sin\theta - gt

y = u\sin\theta \cdot t - \frac{1}{2}gt^2

The resultant speed at any instant $t$ :

v = \sqrt{v_x^2 + v_y^2} = \sqrt{(u\cos\theta)^2 + (u\sin\theta - gt)^2}

Pro tip: Horizontal velocity never changes — gravity acts only vertically. This is why

v_x = u\cos\theta

at every point on the trajectory, including at maximum height and at landing.

3. Key Projectile Formulas

Three quantities completely characterise the trajectory of a ground-launched projectile. You must be able to derive and apply all three.

\text{Time of flight:} \quad T = \frac{2u\sin\theta}{g}

\text{Maximum height:} \quad H = \frac{u^2\sin^2\theta}{2g}

\text{Horizontal range:} \quad R = \frac{u^2\sin 2\theta}{g}

Maximum range occurs when $\sin 2\theta$ is maximum, i.e., $2\theta = 90°$ , so $\theta = 45°$ :

R_{\max} = \frac{u^2}{g} \quad \text{at } \theta = 45°

Complementary angles: Angles $\theta$ and $(90° - \theta)$ give the same range because $\sin 2\theta = \sin(180° - 2\theta)$ . For example, a ball thrown at 30° travels the same horizontal distance as one thrown at 60° (at the same speed), but with different heights and flight times.

NEET tip: Quick check — if range is asked and two angles are given that add to 90°, the range is the same for both. This appears frequently as a conceptual MCQ.

4. Equation of Trajectory

The path traced by a projectile is called its trajectory. To find the equation of the path, eliminate time $t$ from the parametric equations.

From horizontal: $t = \dfrac{x}{u\cos\theta}$ . Substitute into the vertical equation:

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

This is the equation of trajectory. It is of the form $y = Ax - Bx^2$ where $A = \tan\theta$ and $B = \dfrac{g}{2u^2\cos^2\theta}$ are constants for a given launch.

This is the equation of a parabola — the shape of every projectile path in a uniform gravitational field with no air resistance. The parabola opens downward (since the $x^2$ coefficient is negative).

Caution: The trajectory equation assumes launch from the origin (0,0). If the launch point is not the origin, shift the coordinate system accordingly before applying the formula.

Pro tip: In NEET, the trajectory equation is sometimes given in modified form

y = x\tan\theta\left(1 - \dfrac{x}{R}\right)

where R is the range. This form makes it easy to read off the range directly.

5. Projectile: Special Cases

Horizontal projection from height h: Initial vertical velocity = 0, initial horizontal velocity = $u$ .

\text{Time to fall: } T = \sqrt{\frac{2h}{g}}, \quad \text{Range: } R = u \cdot T = u\sqrt{\frac{2h}{g}}

At maximum height: $v_y = 0$ (vertical component vanishes), but $v_x = u\cos\theta$ still acts. So the projectile moves purely horizontally at the peak.

Velocity makes 45° with horizontal when $v_y = v_x$ , i.e., $u\sin\theta - gt = u\cos\theta$ , giving $t = \dfrac{u(\sin\theta - \cos\theta)}{g}$ .

Projectile from a moving platform: Use relative velocity — combine the platform velocity with the projectile velocity vectorially to get the actual velocity in the ground frame.

NEET tip: For horizontal projection, the time of flight depends only on the height

h

and

g

— it is independent of the horizontal speed

u

. A ball thrown horizontally from a cliff and one dropped vertically from the same cliff hit the ground at the same time.

6. Uniform Circular Motion

Uniform circular motion (UCM) is motion along a circular path at constant speed. "Uniform" refers to constant speed, not constant velocity — the direction changes continuously, so velocity is never constant.

Key relations:

\omega = \frac{2\pi}{T} = 2\pi f

v = r\omega

a_c = \frac{v^2}{r} = r\omega^2 = v\omega

F_c = \frac{mv^2}{r} = mr\omega^2 \quad \text{(directed toward center)}

The centripetal acceleration always points toward the center of the circle. It changes the direction of velocity without changing its magnitude. There is no tangential acceleration in UCM — speed is constant.

Caution: Centripetal force is not a new type of force — it is the name given to the net inward force required to maintain circular motion. It is provided by gravity (for satellites), tension (for circular orbits on strings), friction (for cars turning), or normal force (for loop-the-loops).

NEET tip: In UCM, speed is constant but velocity, acceleration, and momentum all change direction continuously. Therefore the kinetic energy is constant, but momentum is not. Centripetal force does no work because it is always perpendicular to velocity.

7. Non-Uniform Circular Motion

When a particle moves in a circle with changing speed, both the magnitude and direction of velocity change. The acceleration now has two perpendicular components.

Tangential acceleration (along the tangent) changes the speed:

a_t = \frac{dv}{dt}

Centripetal (radial) acceleration (along the radius, toward center) changes direction:

a_c = \frac{v^2}{r}

The net acceleration is the vector sum of these two perpendicular components:

a_{\text{net}} = \sqrt{a_t^2 + a_c^2}

\tan\alpha = \frac{a_t}{a_c} \quad \text{(angle with the radius)}

The net force similarly has a radial component (providing centripetal acceleration) and a tangential component (changing speed). A common example is a ball on a string in a vertical circle — at every point both components are present except at the topmost and bottommost points where geometry simplifies the problem.

Pro tip: In a vertical circle under gravity, the critical (minimum) speed at the top of the loop is

v_{\min} = \sqrt{gR}

, found by setting tension = 0 at the top so that gravity alone provides centripetal force.

8. Relative Velocity in 2D

The velocity of object A as observed from object B (the velocity of A relative to B) is obtained by vector subtraction:

\vec{v}_{AB} = \vec{v}_A - \vec{v}_B

Rain-umbrella problem: Rain falls vertically at speed $v_r$ and a man walks horizontally at speed $v_m$ . The velocity of rain relative to the man is $\vec{v}_{\text{rain}} - \vec{v}_{\text{man}}$ , which points obliquely forward-downward. To avoid getting wet, the man must tilt his umbrella in the direction of this relative velocity.

River-crossing problem: A swimmer can swim at speed $u$ in still water; river flows at speed $v$ .

Minimum time crossing: Swim perpendicular to bank. Time $T = d/u$ (d = width). Drift = $vT = vd/u$ .

Minimum drift (shortest path): Swim at angle $\theta = \sin^{-1}(v/u)$ upstream (only possible if $u > v$ ). Drift = 0.

Drift formula (general): $\text{Drift} = (v - u\sin\phi) \cdot \dfrac{d}{u\cos\phi}$ where $\phi$ is angle with perpendicular.

NEET tip: "Minimum time" and "minimum drift" are two different scenarios. For minimum time, always swim straight across (perpendicular). For minimum drift (zero drift), aim upstream — but this is only possible when swimmer speed exceeds river speed.

9. Angular Quantities

Just as linear motion has displacement, velocity, and acceleration, rotational motion has analogous angular quantities. These are not independent — they are connected to linear quantities through the radius $r$ .

Angular displacement: $\theta$ (radians)

Angular velocity: $\omega = \dfrac{d\theta}{dt}$ (rad/s)

Angular acceleration: $\alpha = \dfrac{d\omega}{dt}$ (rad/s²)

Equations of angular kinematics (analogous to $v = u+at$ , $s = ut + \frac{1}{2}at^2$ , $v^2 = u^2 + 2as$ ):

\omega = \omega_0 + \alpha t

\theta = \omega_0 t + \frac{1}{2}\alpha t^2

\omega^2 = \omega_0^2 + 2\alpha\theta

Connecting linear and angular quantities: $v = r\omega$ , $a_t = r\alpha$ , $a_c = r\omega^2 = v^2/r$ .

Pro tip: The structural analogy between linear and angular kinematics is exact: replace

s \to \theta

u \to \omega_0

v \to \omega

a \to \alpha

. Every linear equation has a direct angular counterpart. This analogy saves memorisation time in NEET.

10. NEET Exam Traps for Projectile & Circular Motion

These are the most commonly tested conceptual points and the most common sources of wrong answers in NEET. Study each one carefully.

Trap 1 — Same range for complementary angles: Range is the same for $\theta$ and $(90° - \theta)$ , but time of flight and maximum height are different. If NEET asks "which quantities are the same," only range qualifies — not H or T.

Trap 2 — Centripetal force does NO work: Since centripetal force is always perpendicular to displacement (velocity), it does zero work. Kinetic energy (and thus speed) is constant in UCM. This is a direct consequence of the work-energy theorem.

Trap 3 — Maximum range is at 45° only on flat ground: If the projectile is launched from a height or lands at a different height, the optimal angle shifts below 45°. The formula $R_{\max} = u^2/g$ applies only for ground-to-ground launch on a horizontal surface.

Trap 4 — Speed vs velocity in UCM: In uniform circular motion, speed is constant but velocity is not (direction changes). Therefore momentum is not constant, and the object is accelerating even though its speed is unchanged. Newton's first law is not violated because there is a net force (centripetal).

Trap 5 — At maximum height, only horizontal velocity remains: At the peak of a projectile's path, $v_y = 0$ but $v_x = u\cos\theta \neq 0$ . The projectile is not momentarily at rest (unless it was launched vertically). Acceleration is still $g$ downward at the peak.

NEET tip: In NEET, about 2–3 questions from projectile and circular motion appear every year. Mastering the five traps above and the three key projectile formulas (T, H, R) covers the majority of possible questions from this chapter.

Solved Practice

10 Solved Examples

Open only after you try the question yourself

NEET Exam Trap: At maximum height, a projectile still has horizontal velocity $u\cos\theta$. It is not momentarily at rest. Centripetal force does zero work. Maximum range is at 45° only for flat ground launches.

Example 1. A ball is thrown at 20 m/s at 30° above horizontal. Find the time of flight and maximum height. (g = 10 m/s²)

Vertical component: $u_y = 20\sin 30° = 10$ m/s.

Time of flight: $T = \dfrac{2u_y}{g} = \dfrac{2 \times 10}{10} = 2$ s.

Maximum height: $H = \dfrac{u_y^2}{2g} = \dfrac{100}{20} = 5$ m.

Example 2. A projectile is launched at 45°. What is the ratio of maximum height to range?

At $\theta = 45°$ : $H = \dfrac{u^2\sin^2 45°}{2g} = \dfrac{u^2}{4g}$ and $R = \dfrac{u^2\sin 90°}{g} = \dfrac{u^2}{g}$ .

Ratio $H/R = \dfrac{u^2/4g}{u^2/g} = \dfrac{1}{4}$ .

Example 3. A stone is dropped from a cliff of height 80 m. A ball is thrown horizontally at 15 m/s from the same height. Which hits the ground first? (g = 10 m/s²)

Both have zero initial vertical velocity. Time to fall height $h$ : $T = \sqrt{2h/g} = \sqrt{16} = 4$ s for both.

They hit the ground simultaneously — horizontal speed does not affect time of fall.

Example 4. A particle moves in a circle of radius 2 m at a constant speed of 4 m/s. Find centripetal acceleration and the force if mass = 0.5 kg.

$a_c = v^2/r = 16/2 = 8$ m/s².

$F_c = ma_c = 0.5 \times 8 = 4$ N directed toward the center.

Example 5. Rain falls vertically at 3 m/s. A man walks east at 4 m/s. At what angle should he hold his umbrella?

Velocity of rain relative to man: $\vec{v}_{rel} = \vec{v}_{rain} - \vec{v}_{man} = -3\hat{j} - 4\hat{i}$ (downward and westward).

Angle with vertical: $\tan\theta = 4/3$ , so $\theta = 53°$ tilted forward (east).

Example 6. Find the range of a projectile fired at 60° with speed 20 m/s. (g = 10 m/s²)

$R = \dfrac{u^2\sin 2\theta}{g} = \dfrac{400\sin 120°}{10} = \dfrac{400 \times (\sqrt{3}/2)}{10} = 20\sqrt{3} \approx 34.6$ m.

Example 7. A particle starts from rest and moves in a circle of radius 1 m with angular acceleration α = 2 rad/s². Find the net acceleration after 2 s.

After 2 s: $\omega = \omega_0 + \alpha t = 0 + 2(2) = 4$ rad/s, $v = r\omega = 4$ m/s.

$a_c = v^2/r = 16$ m/s², $a_t = r\alpha = 1 \times 2 = 2$ m/s².

$a_{net} = \sqrt{16^2 + 2^2} = \sqrt{256+4} = \sqrt{260} \approx 16.1$ m/s².

Example 8. For a projectile, prove that the trajectory is parabolic for launch from origin at angle θ.

$x = u\cos\theta \cdot t \Rightarrow t = x/(u\cos\theta)$ .

$y = u\sin\theta \cdot t - \frac{1}{2}gt^2 = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$ .

This is $y = Ax - Bx^2$ form — a downward parabola. QED.

Example 9. A swimmer can swim at 5 m/s in still water. A river flows at 3 m/s. What angle should the swimmer take to cross with zero drift? Find the effective crossing speed.

Zero drift requires $u\sin\phi = v$ where $\phi$ is the upstream angle. $\sin\phi = 3/5$ , so $\phi = 37°$ upstream.

Effective speed across: $v_{eff} = \sqrt{u^2 - v^2} = \sqrt{25-9} = 4$ m/s.

Example 10. At what angle of projection is the range equal to the maximum height?

$R = H \Rightarrow \dfrac{u^2\sin 2\theta}{g} = \dfrac{u^2\sin^2\theta}{2g}$ .

$2\sin\theta\cos\theta = \sin^2\theta/2 \Rightarrow 4\cos\theta = \sin\theta \Rightarrow \tan\theta = 4 \Rightarrow \theta = \tan^{-1}(4) \approx 76°$ .

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