Chapter 4: Motion in a Plane

The world is not one-dimensional. A cricket ball arcs through the air, a stone tied to a string whirls in a horizontal circle, a swimmer crosses a flowing river. To describe such motion we need vectors — quantities with both magnitude and direction — and we need 2D extensions of the kinematic equations of Chapter 3. This chapter sets up the vector toolkit you will use for every chapter after this, and applies it to two iconic problems: projectile motion and uniform circular motion.

Key Concepts

4.1 Scalars and Vectors

• A scalar is a quantity specified completely by a number (with units). Examples: mass, temperature, time, energy, distance, speed.
• A vector is a quantity specified by both a magnitude and a direction. Examples: displacement, velocity, acceleration, force, momentum.

A vector is denoted $\vec{A}$, with magnitude $\vert \vec{A}\vert = A \ge 0$. Two vectors are equal if and only if they have the same magnitude and the same direction — regardless of where they are drawn.

Position vector. In a chosen frame with origin $O$, the position of a particle at $P$ is the vector $\vec{r} = \overrightarrow{OP}$. The displacement in going from $P_1$ to $P_2$ is $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1$ — independent of the path.

4.2 Types of Vectors

• Equal vectors: same magnitude and direction.
• Negative of a vector: $-\vec{A}$ has the same magnitude as $\vec{A}$ but opposite direction.
• Zero / null vector $\vec{0}$: magnitude 0, direction indeterminate. $\vec{A} + (-\vec{A}) = \vec{0}$.
• Unit vector $\hat{A} = \vec{A}/\vert \vec{A}\vert$: magnitude 1, same direction. The Cartesian unit vectors $\hat{i}, \hat{j}, \hat{k}$ point along the $+x, +y, +z$ axes.
• Co-initial vectors: share the same starting point.
• Collinear vectors: lie along the same line (parallel or antiparallel).
• Coplanar vectors: lie in the same plane.

4.3 Vector Addition — Geometric Methods

(a) Triangle law. To add $\vec{A} + \vec{B}$, draw $\vec{A}$, then draw $\vec{B}$ starting from the head of $\vec{A}$. The resultant $\vec{R}$ is drawn from the tail of $\vec{A}$ to the head of $\vec{B}$.

(b) Parallelogram law. If $\vec{A}$ and $\vec{B}$ are drawn from the same point as adjacent sides of a parallelogram, the diagonal from that point is $\vec{R} = \vec{A} + \vec{B}$.

(c) Polygon law. For more than two vectors, place them head-to-tail in any order; the resultant goes from the first tail to the last head.

Properties of vector addition:

• Commutative: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$.
• Associative: $(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C})$.
• Distributive over scalar multiplication: $k(\vec{A} + \vec{B}) = k\vec{A} + k\vec{B}$.

Magnitude and direction of the resultant of two vectors at angle $\theta$ (parallelogram law): $R = \sqrt{A^2 + B^2 + 2AB \cos\theta},$ $\tan\alpha = \frac{B \sin\theta}{A + B \cos\theta},$ where $\alpha$ is the angle that $\vec{R}$ makes with $\vec{A}$.

Special cases:

• $\theta = 0$: $R_{\max} = A + B$.
• $\theta = 180^\circ$: $R_{\min} = \vert A - B\vert$.
• $\theta = 90^\circ$: $R = \sqrt{A^2 + B^2}$.

For any two vectors of magnitudes $A$ and $B$, the magnitude of the resultant satisfies $\vert A - B\vert \le R \le A + B$.

4.4 Subtraction, Multiplication by a Scalar

• Subtraction: $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$.
• Multiplication by a positive scalar $k$: same direction, magnitude scaled by $k$.
• Multiplication by a negative scalar: direction reverses.

4.5 Resolution of a Vector — Components

Any vector in a plane can be expressed as a sum of two non-collinear components. The most useful choice is the rectangular (Cartesian) components: $\vec{A} = A_x \hat{i} + A_y \hat{j},$ where $A_x = A \cos\theta$, $A_y = A \sin\theta$ and $\theta$ is the angle with the $+x$-axis. Then: $A = \sqrt{A_x^2 + A_y^2}, \qquad \tan\theta = \frac{A_y}{A_x}.$

In 3D: $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$, $A = \sqrt{A_x^2 + A_y^2 + A_z^2}$.

Direction cosines: $\cos\alpha = A_x/A$, $\cos\beta = A_y/A$, $\cos\gamma = A_z/A$ — satisfy $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.

Analytical addition. If $\vec{R} = \vec{A} + \vec{B}$: $R_x = A_x + B_x, \qquad R_y = A_y + B_y.$ Then magnitude and direction follow from the formulae above. This is by far the most reliable computational approach.

4.6 Scalar (Dot) Product

$\vec{A} \cdot \vec{B} = AB \cos\theta = A_x B_x + A_y B_y + A_z B_z.$

Properties:

• Scalar result (no direction).
• Commutative: $\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$.
• Distributive: $\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}$.
• $\vec{A} \cdot \vec{A} = A^2$.
• $\vec{A} \cdot \vec{B} = 0$ iff the vectors are perpendicular (or one is zero).
• $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$; $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$.

Geometric meaning. $\vec{A} \cdot \vec{B}$ is the magnitude of $\vec{A}$ times the projection of $\vec{B}$ onto $\vec{A}$ (with sign).

Physical applications:

• Work done: $W = \vec{F} \cdot \vec{d}$.
• Power: $P = \vec{F} \cdot \vec{v}$.
• Flux of a field through an area: $\Phi = \vec{E} \cdot \vec{A}$.

4.7 Vector (Cross) Product

$\vec{A} \times \vec{B} = (AB \sin\theta)\,\hat{n},$ where $\hat{n}$ is a unit vector perpendicular to the plane of $\vec{A}$ and $\vec{B}$, in the direction given by the right-hand rule (curl fingers from $\vec{A}$ to $\vec{B}$; thumb gives $\hat{n}$).

Component form: $\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} = (A_y B_z - A_z B_y) \hat{i} + (A_z B_x - A_x B_z) \hat{j} + (A_x B_y - A_y B_x) \hat{k}.$

Properties:

• Vector result, perpendicular to both inputs.
• Anti-commutative: $\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}$.
• Distributive: $\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}$.
• $\vec{A} \times \vec{A} = \vec{0}$.
• $\vec{A} \times \vec{B} = \vec{0}$ iff vectors are parallel or antiparallel (or one is zero).
• $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$; cyclic.

Geometric meaning. $\vert \vec{A} \times \vec{B}\vert$ equals the area of the parallelogram spanned by $\vec{A}$ and $\vec{B}$; half this is the area of the triangle.

Physical applications:

• Torque: $\vec{\tau} = \vec{r} \times \vec{F}$.
• Angular momentum: $\vec{L} = \vec{r} \times \vec{p}$.
• Magnetic force on a moving charge: $\vec{F} = q \vec{v} \times \vec{B}$.

4.8 Position, Velocity, Acceleration in 2D

For a particle with position vector $\vec{r}(t) = x(t) \hat{i} + y(t) \hat{j}$: $\vec{v} = \frac{d\vec{r}}{dt} = \dot{x} \hat{i} + \dot{y} \hat{j}, \qquad \vec{a} = \frac{d\vec{v}}{dt} = \ddot{x} \hat{i} + \ddot{y} \hat{j}.$ In 2D, $\vec{v}$ is tangent to the path at every point.

Equations of motion for constant $\vec{a}$ — same as 1D, but vector-valued: $\vec{v} = \vec{u} + \vec{a} t, \quad \vec{r} = \vec{r}_0 + \vec{u} t + \tfrac{1}{2} \vec{a} t^2.$ Key idea: components along perpendicular axes are independent. Choose axes wisely and solve as two 1D problems.

4.9 Projectile Motion

A projectile is an object given an initial velocity and then moving freely under gravity (no air resistance, no thrust). The acceleration is $\vec{g} = -g \hat{j}$ throughout.

Set up axes at the launch point with $x$ horizontal (direction of horizontal velocity), $y$ vertical upward. Initial velocity at angle $\theta$ to the horizontal: $u_x = u\cos\theta$, $u_y = u\sin\theta$.

Independent component motions:

• Horizontal: $a_x = 0$, $v_x = u\cos\theta$ (constant), $x = u\cos\theta \cdot t$.
• Vertical: $a_y = -g$, $v_y = u\sin\theta - g t$, $y = u\sin\theta \cdot t - \tfrac{1}{2} g t^2$.

Equation of trajectory. Eliminate $t$ using $t = x/(u\cos\theta)$: $y = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta}.$ This is a downward parabola — the signature curve of projectile motion. Galileo first proved this geometrically in 1638.

Key quantities (projectile launched from ground, returning to ground):

• Time of flight: vertical displacement zero $\Rightarrow t = 0$ or $t = T$ with $T = \frac{2 u \sin\theta}{g}.$
• Maximum height: $H = \frac{u^2 \sin^2\theta}{2g}.$
• Horizontal range: $R = u\cos\theta \cdot T = \frac{u^2 \sin 2\theta}{g}.$
• Maximum range at $\theta = 45^\circ$: $R_{\max} = u^2 / g$.
• Velocity at any time: $\vec{v}(t) = u\cos\theta\, \hat{i} + (u\sin\theta - gt)\, \hat{j}$. Speed $= \sqrt{v_x^2 + v_y^2}$. At the peak, $v_y = 0$ and $v = v_x = u\cos\theta$.

Symmetric properties (launch and landing at same height):

• Same speed of projection and landing.
• Same magnitude of velocity but reversed $y$-component.
• Equal range achieved at two complementary angles $\theta$ and $90^\circ - \theta$ (because $\sin 2\theta = \sin(180^\circ - 2\theta)$). Times of flight differ, but ranges coincide.

Derivation of $H$. At peak $v_y = 0 = u\sin\theta - g t_H \Rightarrow t_H = u\sin\theta / g$. $H = u\sin\theta \cdot t_H - \tfrac{1}{2} g t_H^2 = u^2 \sin^2\theta / (2g)$.

Derivation of $R$. $R = u\cos\theta \cdot T = u\cos\theta \cdot 2u\sin\theta/g = u^2 \sin 2\theta / g$ using $2\sin\theta\cos\theta = \sin 2\theta$.

Projectile from a height $h$. A particle projected horizontally with speed $u$ from height $h$ above ground:

• Time of flight: $t = \sqrt{2h/g}$.
• Horizontal range: $R = u\sqrt{2h/g}$.
• Velocity on hitting ground: $\vec{v} = u\hat{i} - \sqrt{2gh}\,\hat{j}$; speed $= \sqrt{u^2 + 2gh}$.

4.10 Uniform Circular Motion

A particle moving in a circle of radius $r$ with constant speed $v$ is in uniform circular motion. The velocity is tangential, but it is constantly changing direction — so there is a non-zero acceleration even though the speed is constant.

Angular variables.

• Angular displacement $\theta$ (in radians).
• Angular velocity $\omega = d\theta / dt$ (rad/s).
• Period $T = 2\pi / \omega$ (s).
• Frequency $f = 1/T = \omega/(2\pi)$ (Hz).
• Linear-angular relation: $v = \omega r$.

Centripetal acceleration. The acceleration in uniform circular motion is directed toward the center (radially inward) and has magnitude $a_c = \frac{v^2}{r} = \omega^2 r = v \omega.$

Derivation (geometric). In a small time $\Delta t$ the position vector rotates by $\Delta\theta = \omega \Delta t$. The velocity vector also rotates by $\Delta\theta$ (it stays perpendicular to $\vec{r}$). The two velocity vectors $\vec{v}(t)$ and $\vec{v}(t + \Delta t)$, both of magnitude $v$, form an isoceles triangle with apex angle $\Delta\theta$. For small $\Delta\theta$: $|\Delta\vec{v}| \approx v\,\Delta\theta = v \omega \Delta t.$ Direction of $\Delta\vec{v}$: in the limit $\Delta t \to 0$, it points radially inward (toward the center). So $|\vec{a}| = \lim_{\Delta t \to 0} \frac{|\Delta\vec{v}|}{\Delta t} = v\omega = \frac{v^2}{r}.$

Derivation (calculus). Write $\vec{r}(t) = r(\cos\omega t\,\hat{i} + \sin\omega t\,\hat{j})$. Differentiate: $\vec{v} = r\omega(-\sin\omega t\,\hat{i} + \cos\omega t\,\hat{j}), \quad |\vec{v}| = r\omega = v.$ $\vec{a} = r\omega^2(-\cos\omega t\,\hat{i} - \sin\omega t\,\hat{j}) = -\omega^2 \vec{r}.$ Magnitude $\omega^2 r$, direction opposite to $\vec{r}$ — i.e., toward the center. $\checkmark$

Centripetal force. Required for circular motion: $F = m a_c = m v^2 / r = m \omega^2 r$. This force can be provided by tension (a stone on a string), gravity (a satellite in orbit), friction (a car turning), normal force (a roller coaster loop), or a combination.

Non-uniform circular motion (out of NCERT scope but useful): if the speed also changes, there is an additional tangential acceleration $a_t = dv/dt$. The total acceleration is $\vec{a} = \vec{a}_c + \vec{a}_t$ with $\vert \vec{a}\vert = \sqrt{a_c^2 + a_t^2}$.

4.11 Relative Velocity in Two Dimensions

For two particles $A$ and $B$ with velocities $\vec{v}_A$ and $\vec{v}_B$ in a common frame, the velocity of $A$ relative to $B$ is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B.$ This is a vector equation. Magnitude: $|\vec{v}_{AB}| = \sqrt{v_A^2 + v_B^2 - 2 v_A v_B \cos\theta},$ where $\theta$ is the angle between $\vec{v}_A$ and $\vec{v}_B$.

(a) Rain-and-man problem. Rain falls vertically with speed $v_r$ relative to the ground. A man walks with horizontal speed $v_m$. The rain relative to the man has velocity $\vec{v}_{rm} = \vec{v}_r - \vec{v}_m = -v_m \hat{i} - v_r \hat{j}.$ The man should tilt his umbrella by an angle $\alpha = \tan^{-1}(v_m/v_r)$ from the vertical, in the direction of his motion.

(b) River-and-boat problem. A boat with speed $v_b$ (relative to water) crosses a river of width $d$ with current $v_c$ (river velocity relative to ground).

• Shortest time: point the boat perpendicular to the banks. Time $= d/v_b$. Downstream drift $= v_c \cdot d/v_b$. Resultant velocity $\sqrt{v_b^2 + v_c^2}$.
• Shortest path (straight across): the boat must aim upstream at angle $\theta$ such that $v_b \sin\theta = v_c$, i.e., $\sin\theta = v_c/v_b$ (requires $v_b > v_c$). Effective speed across $= \sqrt{v_b^2 - v_c^2}$. Time $= d/\sqrt{v_b^2 - v_c^2}$.

(c) Aircraft and wind. Same logic: ground velocity = airspeed (relative to wind) + wind velocity (relative to ground).

Worked Examples

Example 4.1 — Resultant of two vectors. Two forces of magnitudes $6$ N and $8$ N act on a particle at $60^\circ$ to each other. Find the magnitude and direction of the resultant.

Solution. $R = \sqrt{6^2 + 8^2 + 2\cdot 6\cdot 8 \cos 60^\circ} = \sqrt{36 + 64 + 48} = \sqrt{148} \approx 12.17$ N. $\tan\alpha = (8 \sin 60^\circ) / (6 + 8\cos 60^\circ) = (8 \cdot 0.866) / (6 + 4) = 6.928/10 = 0.693$, so $\alpha \approx 34.7^\circ$ from the $6$ N force.

Example 4.2 — Dot and cross. $\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$, $\vec{B} = \hat{i} - 2\hat{j} + 3\hat{k}$. Find $\vec{A} \cdot \vec{B}$, $\vec{A} \times \vec{B}$, the angle between them, and a unit vector perpendicular to both.

Solution. $\vec{A} \cdot \vec{B} = 2(1) + 3(-2) + (-1)(3) = 2 - 6 - 3 = -7$. $\vert \vec{A}\vert = \sqrt{4+9+1} = \sqrt{14}$, $\vert \vec{B}\vert = \sqrt{1+4+9} = \sqrt{14}$. $\cos\theta = -7/14 = -0.5 \Rightarrow \theta = 120^\circ$. $\vec{A} \times \vec{B} = (3\cdot 3 - (-1)(-2))\hat{i} - (2\cdot 3 - (-1)(1))\hat{j} + (2(-2) - 3(1))\hat{k} = (9 - 2)\hat{i} - (6 + 1)\hat{j} + (-4 - 3)\hat{k} = 7\hat{i} - 7\hat{j} - 7\hat{k}$. $\vert \vec{A} \times \vec{B}\vert = 7\sqrt{3}$. Unit vector: $\hat{n} = (1/\sqrt{3})(\hat{i} - \hat{j} - \hat{k})$.

Cross-check: $\vert \vec{A} \times \vec{B}\vert = AB\sin\theta = 14 \cdot \sin 120^\circ = 14 \cdot \sqrt{3}/2 = 7\sqrt{3}$. $\checkmark$

Example 4.3 — Projectile basics. A ball is thrown with speed $20$ m/s at $30^\circ$ above the horizontal from the ground. (Take $g = 10$ m/s$^2$.) Find time of flight, maximum height, range, and the velocity vector at $t = 1$ s.

Solution. $u_x = 20 \cos 30^\circ = 10\sqrt{3}\approx 17.32$ m/s. $u_y = 20\sin 30^\circ = 10$ m/s. $T = 2 u_y / g = 20/10 = 2$ s. $H = u_y^2 / (2g) = 100/20 = 5$ m. $R = u_x \cdot T = 17.32 \cdot 2 = 34.64$ m. (Cross-check: $R = u^2 \sin 60^\circ / g = 400 \cdot 0.866/10 = 34.64$ m. $\checkmark$) At $t = 1$ s: $v_x = 17.32$ m/s, $v_y = 10 - 10\cdot 1 = 0$ m/s. So at the peak: $\vec{v} = 17.32 \hat{i}$ m/s, speed $17.32$ m/s.

Example 4.4 — Equation of trajectory. For the previous launch, write $y$ as a function of $x$.

Solution. $y = x \tan 30^\circ - g x^2/(2 u^2 \cos^2 30^\circ) = x/\sqrt{3} - 10 x^2 / (2 \cdot 400 \cdot 3/4) = x/\sqrt{3} - x^2/60$. Parabola opening downward.

Example 4.5 — Projectile from a height. A stone is projected horizontally with speed $20$ m/s from the top of a $45$ m tall cliff. (a) Time to hit the ground; (b) horizontal distance from the foot of the cliff; (c) velocity on impact.

Solution. (a) $h = \tfrac{1}{2} g t^2 \Rightarrow t = \sqrt{2 \cdot 45 / 10} = 3$ s. (b) $x = u t = 20 \cdot 3 = 60$ m. (c) $v_x = 20$ m/s, $v_y = -g t = -30$ m/s. Speed $= \sqrt{400 + 900} = \sqrt{1300} \approx 36.06$ m/s. Angle below horizontal $\tan^{-1}(30/20) = 56.3^\circ$.

Example 4.6 — Centripetal acceleration. A satellite orbits Earth in a circle of radius $7000$ km with a period of $5800$ s. Find its speed and centripetal acceleration.

Solution. $v = 2\pi r / T = 2\pi \cdot 7 \times 10^6 / 5800 \approx 7580$ m/s. $a_c = v^2/r = (7580)^2 / 7 \times 10^6 \approx 5.75 \times 10^7 / 7 \times 10^6 \approx 8.21$ m/s$^2$. This is consistent with the value of $g$ at this altitude (slightly less than at the surface).

Example 4.7 — River and boat. A river is $200$ m wide. A boat can move at $4$ m/s in still water. The river flows at $3$ m/s. (a) Minimum time to cross? Where does the boat land? (b) Boat aimed to land directly across — at what angle, and how long does it take?

Solution. (a) Aim perpendicular to bank. $t_{\min} = 200/4 = 50$ s. Downstream drift $= 3 \cdot 50 = 150$ m. (Resultant ground speed $\sqrt{16 + 9} = 5$ m/s.) (b) Aim upstream by $\theta$ such that $4\sin\theta = 3 \Rightarrow \sin\theta = 0.75 \Rightarrow \theta = 48.6^\circ$ upstream of perpendicular. Effective across speed $= 4\cos\theta = \sqrt{16 - 9} = \sqrt{7} \approx 2.646$ m/s. Time $= 200/2.646 \approx 75.6$ s.

Example 4.8 — Rain and umbrella. Rain falls vertically at $5$ m/s. A man cycles east at $5$ m/s. In which direction should he tilt his umbrella?

Solution. Rain relative to man: $\vec{v}_{rm} = -5 \hat{i} - 5 \hat{j}$ m/s. Magnitude $5\sqrt{2}$ m/s. Angle with vertical $= \tan^{-1}(5/5) = 45^\circ$ toward east (the direction of motion). He should tilt the umbrella $45^\circ$ forward.

Example 4.9 — Two equal ranges. A projectile has range $40$ m when fired at $30^\circ$. Find the other angle (with same speed) giving the same range, and verify.

Solution. Range depends on $\sin 2\theta$, which is symmetric about $\theta = 45^\circ$. The complementary angle is $60^\circ$. $\sin 60^\circ = \sin 120^\circ$, so both give the same $R$. Time of flight at $60^\circ$ is larger by factor $\tan 60^\circ / \tan 30^\circ = 3$.

Common Traps

• Adding magnitudes instead of vectors. $\vert \vec{A} + \vec{B}\vert \ne A + B$ in general.
• Forgetting the angle convention. "Angle between two forces" usually means the angle when their tails are placed together — not when they are placed head-to-tail.
• Believing distance traveled equals magnitude of displacement. Only true for straight-line, unidirectional motion.
• At the peak of a projectile, "everything is zero." Velocity is not zero — only its vertical component is. Horizontal component remains $u\cos\theta$.
• Air resistance ignored. Standard projectile formulas assume vacuum or "no air drag." Real-world projectiles deviate.
• Confusing $\sin 2\theta$ and $2\sin\theta\cos\theta$. They are equal — but a common source of arithmetic slips.
• Centripetal force is a separate "extra" force. It is just the net force pointing toward the center; the physical agents are tension, gravity, friction, etc.
• Centrifugal force in inertial frames. Centrifugal force does not exist in an inertial frame. It is a pseudo-force that appears only in a rotating (non-inertial) frame.
• At the peak, acceleration is zero. No — acceleration is $g$ downward throughout. Velocity changes direction here.
• Dot product of perpendicular vectors is not always zero. Yes it is — if the vectors are non-zero.
• Cross product magnitude formula sign. $\vert \vec{A} \times \vec{B}\vert = AB\vert \sin\theta\vert$ — always positive. The cross product as a vector has direction from the right-hand rule.
• Using complementary-angle range formula off the ground. If launch and landing heights differ, complementary angles do not give equal ranges.
• Range formula valid only for landing at launch height. For inclined surfaces or different landing heights, derive afresh.
• Mixing speed and angular speed. $v = \omega r$ links them — units must match (rad/s, m, m/s).
• In river-crossing, confusing "across" velocity with "ground" velocity. Time to cross = (width)/(component of velocity perpendicular to banks). Use the perpendicular component, not the resultant.
• Using $T = 2u/g$ for projectile that lands at different height. That is the symmetric-throw formula; in general, solve the quadratic for $y(t) = h_{\text{land}}$.

Quick Recap

• Scalars have only magnitude; vectors have magnitude and direction. Position, displacement, velocity, acceleration, force, momentum, torque are vectors.
• Resultant of $\vec{A}$ and $\vec{B}$ at angle $\theta$: $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$, $\tan\alpha = B\sin\theta/(A + B\cos\theta)$.
• $\vert A - B\vert \le R \le A + B$ for any two vectors.
• Rectangular components: $\vec{A} = A\cos\theta\, \hat{i} + A\sin\theta\, \hat{j}$. Add component-wise.
• Dot product: $\vec{A} \cdot \vec{B} = AB\cos\theta$, scalar; zero when perpendicular. Used for work, power, flux.
• Cross product: $\vec{A} \times \vec{B} = AB\sin\theta\, \hat{n}$, vector, right-hand rule; zero when parallel. Used for torque, angular momentum, magnetic force.
• For constant $\vec{a}$ in 2D: components along $x$ and $y$ are independent 1D problems.
• Projectile (launched from ground):
  • Trajectory $y = x\tan\theta - g x^2/(2 u^2 \cos^2\theta)$ — parabola.
  • $T = 2u\sin\theta/g$, $H = u^2\sin^2\theta/(2g)$, $R = u^2\sin 2\theta/g$.
  • $R_{\max} = u^2/g$ at $\theta = 45^\circ$.
  • Complementary angles $\theta$ and $90^\circ - \theta$ give the same range.
• Uniform circular motion: $v = \omega r$; $a_c = v^2/r = \omega^2 r$ directed radially inward.
• $T = 2\pi/\omega$, $f = 1/T$.
• Relative velocity in 2D: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$.
• River-crossing: minimum time vs minimum drift — pick axes properly.
• Rain-umbrella: tilt the umbrella in the direction of motion by $\tan^{-1}(v_m/v_r)$.

Formula Summary

With this 2D toolkit, you can now describe most everyday motion — and you have the algebra needed to step into Newton's laws of motion in the next chapter.