The instantaneous velocity is always tangent to the path.
In 2-D, acceleration can change the direction of velocity even at constant speed
(uniform circular motion) — impossible in 1-D.
The angle between $\vec{v}$ and $\vec{a}$ decides the motion: 0° → speeding up along a straight line;
180° → slowing; 90° → circular; any other fixed angle with constant $\vec{a}$ → parabolic (projectile).
Figure 2b
Velocity is tangent to the path; the v–a angle decides the shape of motion
The four cases of the bullet list above, drawn: green is v, red is a.
1.1 Motion with Constant Acceleration in a Plane
The Master Principle
(NCERT's most important statement in this chapter): Motion in a plane is
two independent 1-D motions along perpendicular axes. Each axis obeys its own set of
kinematic equations:
$$\vec{v} = \vec{u} + \vec{a}t \quad\Longleftrightarrow\quad
\begin{cases} v_x = u_x + a_x t \\ v_y = u_y + a_y t \end{cases}$$
$$\vec{r} = \vec{u}t + \tfrac12 \vec{a}t^2 \quad\Longleftrightarrow\quad
\begin{cases} x = u_x t + \tfrac12 a_x t^2 \\ y = u_y t + \tfrac12 a_y t^2 \end{cases}$$
Time t is the only quantity common to both axes. This single
idea solves all of projectile motion.
✍ IN-TEXT PRACTICE 1.1 — NCERT / JEE (Calculus in 2-D)
Q. The position of a particle is $\vec{r} = 3t\hat{i} + 2t^{2}\hat{j}$ (metres, seconds).
Find velocity and acceleration at t = 2 s, and the speed at that instant.
Velocity: $\vec{v} = \dfrac{d\vec{r}}{dt} = 3\hat{i} +
4t\hat{j}$; at t = 2 s: $\vec{v} = 3\hat{i} + 8\hat{j}$ m/s
Acceleration: $\vec{a} = \dfrac{d\vec{v}}{dt} = 4\hat{j}$ m/s²
(constant, along y)
Aim upstream at angle θ with the ⊥ so that $v_b\sin\theta = v_r$
$\sin\theta = \dfrac{v_r}{v_b}$ (possible only if $v_b > v_r$); crossing speed =
$\sqrt{v_b^2 - v_r^2}$; $t = \dfrac{d}{\sqrt{v_b^2 - v_r^2}}$
⭐ EXAM TRAP
If $v_r > v_b$, zero drift is impossible —
the boat can only minimise drift, not eliminate it. Also note: crossing time depends only on the
perpendicular component of boat velocity, never on the river speed.
2.2 Rain–Man Problems
Method
Velocity of rain relative to man: $\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$. The man must hold his umbrella
along $\vec{v}_{rm}$.
Standard case: rain falls vertically at $v_r$; man walks
horizontally at $v_m$. Then $\vec{v}_{rm}$ makes angle θ with the vertical where
$$\tan\theta = \frac{v_m}{v_r}$$
— tilt the umbrella forward (in the direction of walking) by θ. Apparent speed of rain
$= \sqrt{v_r^2 + v_m^2}$.
Figure 3
River–boat (both strategies, animated) and the rain–man vector triangle
The blue boat aims straight and drifts; the green boat aims upstream and crosses straight. Right: hold the umbrella along v_rm.
✍ IN-TEXT PRACTICE 2.1 — NEET / JEE (River–Boat)
Q. A boat can move at 5 m/s in still water. It crosses a 100 m wide river flowing at 3
m/s. Find (a) the minimum crossing time and the drift in that case, (b) the crossing time for the
shortest path.
(b) Shortest path (zero drift): effective speed $=
\sqrt{5^2-3^2} = 4$ m/s → $t = 100/4 = \mathbf{25 \text{ s}}$ (aim upstream at $\sin\theta = 3/5
\Rightarrow \theta = 37°$ from the perpendicular)
✍ IN-TEXT PRACTICE 2.2 — Boards / NDA (Rain–Man)
Q. Rain falls vertically at 4 m/s. A man walks at 3 m/s on a straight road. At what angle
from the vertical must he hold his umbrella, and what is the apparent speed of the rain?
Angle: $\tan\theta = v_m/v_r = 3/4 \Rightarrow \theta =
\mathbf{37°}$ from the vertical, tilted forward
where $\vec{v}_{p/g}$ is velocity of plane w.r.t ground (actual track), $\vec{v}_{p/a}$ is velocity of plane w.r.t still air (heading/steering direction & airspeed), and $\vec{v}_w$ is wind velocity.
To fly directly from point A to point B in the presence of crosswind, the plane must head at an angle upstream into the wind so that the resultant $\vec{v}_{p/g}$ lies along the AB line.
Round-trip time across a wind of speed $v_w$ perpendicular to the track: $T = \dfrac{2d}{\sqrt{v_{p/a}^2 - v_w^2}}$.
Round-trip time along a wind of speed $v_w$ parallel to the track: $T = \dfrac{d}{v_{p/a} - v_w} + \dfrac{d}{v_{p/a} + v_w} = \dfrac{2d\,v_{p/a}}{v_{p/a}^2 - v_w^2}$.
3. Projectile Motion (Ground-to-Ground) — Complete Theory
A projectile is a body thrown with some initial velocity that then moves under gravity
alone (air resistance neglected). Its path is a parabola. Throughout the flight:
$a_x = 0$ and $a_y = -g$ (taking up as +).
Setup
Projectile fired from the ground with speed u at angle θ with the horizontal.
Components: $u_x = u\cos\theta$ (never changes) and $u_y = u\sin\theta$ (changes due to gravity). The
horizontal motion is uniform; the vertical motion is exactly a "ball thrown up" from the 1-D chapter.
The two are linked only by time.
3.1 Equation of Trajectory (Derivation)
Derivation. Horizontal: $x = (u\cos\theta)t \Rightarrow t = \dfrac{x}{u\cos\theta}$.
Substitute in vertical: $y = (u\sin\theta)t - \tfrac12 gt^2$:
This is of the form $y = ax - bx^2$ — a parabola. Alternative
exam-favourite form: $y = x\tan\theta\left(1 - \dfrac{x}{R}\right)$ where R is the range.
3.2 Time of Flight, Maximum Height & Range (Derivations)
Time of flight T: vertical displacement = 0 at landing:
$0 = (u\sin\theta)T - \tfrac12 gT^2$
Maximum height H: at the top $v_y = 0$: use $v_y^2 = u_y^2 - 2gH$
Range R: $R = u_x \times T = u\cos\theta \cdot
\dfrac{2u\sin\theta}{g}$, then use $2\sin\theta\cos\theta = \sin 2\theta$.
The Big Three (Must Memorise)
$$T = \frac{2u\sin\theta}{g} \qquad H = \frac{u^{2}\sin^{2}\theta}{2g} \qquad R =
\frac{u^{2}\sin 2\theta}{g}$$
3.3 Key Results & Relations (Complete List)
Maximum range: at θ = 45°, $R_{max} = \dfrac{u^2}{g}$, and then $H =
\dfrac{R_{max}}{4}$.
Complementary angles: θ and (90° − θ) give the same range (since
$\sin 2\theta = \sin(180° - 2\theta)$), but different H and T. If $H_1, H_2$ are their heights:
$R = 4\sqrt{H_1H_2}$ and $T_1T_2 = \dfrac{2R}{g}$.
Relation between R and H: $\dfrac{R}{H} = \dfrac{4}{\tan\theta}$ ⟹ $\tan\theta =
\dfrac{4H}{R}$. If R = 4H exactly, θ = 45°; if H = R, tan θ = 4.
Velocity at time t: $v_x = u\cos\theta$, $v_y = u\sin\theta - gt$; speed $=
\sqrt{v_x^2+v_y^2}$; direction $\tan\alpha = v_y/v_x$.
At the top: velocity $= u\cos\theta$ (horizontal, minimum speed of flight),
acceleration is still g downward, and $\vec{v} \perp \vec{a}$ there.
Speed symmetry: speed (and angle magnitude) at the same height is equal going up and
coming down; landing speed = launch speed u at angle −θ.
Change in velocity over the whole flight: $\Delta\vec{v} = \vec{v}_{land} -
\vec{v}_{launch} = -gT\hat{j}$ (magnitude $2u\sin\theta$, vertically downward). Change in momentum $=
2mu\sin\theta$ downward.
Projectile motion is a motion with constant acceleration but continuously changing
velocity (direction changes every instant).
Kinetic energy at the top $= \tfrac12 m(u\cos\theta)^2 = KE_{launch}\cos^2\theta$ (a NEET favourite).
Figure 4
Ground-to-ground projectile with all parameters annotated
Physics-exact animation: the green horizontal-velocity arrow stays constant while the orange vertical component shrinks linearly, vanishes at the peak, and regrows downward; dashed grey twins show the equal-range complementary angles.
✍ IN-TEXT PRACTICE 3.1 — Boards / NEET (Direct formulas)
Q. A ball is projected at 40 m/s at 30° above the horizontal (g = 10 m/s²). Find T, H and
R.
Answer: $t = \dfrac{u}{g\sin\theta}$ (possible only if this ≤ T,
i.e. θ ≥ 45°... check in each problem — a JEE-level subtlety)
4. Projectile from a Height (Horizontal & Angular Projection)
4.1 Horizontal Projectile from Height h
Body thrown horizontally with speed u from height h (e.g., a ball rolled off a table, a packet dropped from
a plane... careful: a dropped packet shares the plane's horizontal velocity!).
Horizontal Projectile Results
$$\text{Time of fall: } t = \sqrt{\frac{2h}{g}} \quad (\text{independent of } u!)$$
$$\text{Horizontal range: } R = u\sqrt{\frac{2h}{g}}$$
$$\text{Landing speed: } v = \sqrt{u^{2} + 2gh}, \qquad \tan\alpha = \frac{gt}{u}$$
$$\text{Trajectory: } y = \frac{gx^{2}}{2u^{2}} \ (\text{measuring y downward})$$
⭐ THE MONKEY-AND-BULLET IDEA (HCV)
A ball dropped from rest and a ball thrown horizontally from
the same height hit the ground at the same time — vertical motion is completely
independent of horizontal velocity. This is the single most repeated conceptual MCQ from this topic in
NEET and NDA.
4.2 Projection at an Angle from a Tower
Thrown from height h with speed u at angle θ above horizontal. Take up = +, origin at the top. Landing when
vertical displacement = −h:
$$-h = (u\sin\theta)t - \tfrac12 gt^{2}$$
Solve the quadratic for t; then range $= (u\cos\theta)t$. Thrown at θ below the horizontal → just
replace $u\sin\theta$ with $-u\sin\theta$. Landing speed always comes fastest from energy form: $v =
\sqrt{u^2 + 2gh}$ — independent of the angle of projection (JEE favourite).
Figure 5b
Projection at an angle from a tower: the −h quadratic and the angle-independent landing speed
Two launch angles (solid blue, dashed grey) from the same tower land at different points and times — but with identical landing speed √(u² + 2gh).
Figure 5
Horizontal projectile from a height, with the dropped-vs-thrown comparison
Left: constant vₓ, growing v_y. Right inset: the orange (dropped) and blue (thrown) balls strike the ground together, every loop.
✍ IN-TEXT PRACTICE 4.1 — Boards / NDA (Horizontal projectile)
Q. A ball is thrown horizontally at 15 m/s from the top of a 20 m tower (g = 10 m/s²).
Find the time of flight, the horizontal distance covered, and the landing speed.
✍ IN-TEXT PRACTICE 4.2 — JEE Main (Packet from an aeroplane)
Q. An aeroplane flying horizontally at 100 m/s at a height of 500 m drops a packet. How
far ahead of the target point (horizontally) must it be released, and where is the plane when the
packet lands? (g = 10 m/s²)
Key idea: the packet starts with the plane's horizontal velocity
(u = 100 m/s), so it is a horizontal projectile.
Time: $t = \sqrt{2\times500/10} = \mathbf{10 \text{ s}}$;
release distance $= 100\times10 = \mathbf{1000 \text{ m}}$ before the target.
Plane's position at landing: the plane is directly above
the packet (both kept the same horizontal velocity) — classic conceptual finish.
5. Uniform Circular Motion (UCM)
A particle moving on a circle of radius r with constant speed is in uniform circular
motion. Speed is constant, but the direction of velocity changes continuously — so there
IS acceleration, directed towards the centre.
5.1 Angular Variables
Quantity
Definition & Formula
SI Unit
Angular displacement θ
Angle swept at the centre; $\theta = \dfrac{\text{arc}}{\text{radius}} = \dfrac{s}{r}$
Time for one revolution; $T = \dfrac{2\pi r}{v} = \dfrac{1}{f}$
second
Frequency f
Revolutions per second; $f = 1/T$
hertz (Hz) / rps
Linear–Angular Link
$$v = \omega r \qquad (\text{vector form: } \vec{v} = \vec{\omega}\times\vec{r})$$
5.2 Centripetal Acceleration (Derivation)
Geometric derivation (NCERT). In a small time Δt the particle turns by Δθ. Velocity
vectors at the two instants have equal magnitude v but differ in direction by Δθ. The velocity triangle
is similar to the position triangle, so
In UCM: $\vec{a} \perp \vec{v}$ always; $\vec{a}$ constantly changes direction — so UCM is accelerated
motion with constant magnitude of acceleration but not constant acceleration vector.
Work done by the centripetal force in UCM is zero (force ⊥ displacement) — speed and
KE stay constant.
Non-uniform circular motion: speed also changes → add a tangential acceleration $a_t =
\dfrac{dv}{dt}$; net acceleration $a = \sqrt{a_c^{2} + a_t^{2}}$, at angle $\tan\phi = a_t/a_c$ with
the radius.
UCM: revolving particle with tangential v and radial a꜀, plus the derivation triangle
Watch the green arrow stay tangent and the red arrow stay centre-pointing as the particle revolves — that IS uniform circular motion. Faint ghost shows a second instant for print.
✍ IN-TEXT PRACTICE 5.1 — NEET / Boards (Direct UCM)
Q. A particle completes 7 revolutions in 44 s on a circle of radius 10 cm. Find its
angular speed, linear speed and centripetal acceleration. (Take π = 22/7)
Time period: $T = 44/7$ s → $\omega = \dfrac{2\pi}{T} =
\dfrac{2\times(22/7)}{44/7} = \mathbf{1 \text{ rad/s}}$
Linear speed: $v = \omega r = 1 \times 0.10 = \mathbf{0.1 \text{
m/s}}$
Centripetal acceleration: $a_c = \omega^2 r = 1 \times 0.10 =
\mathbf{0.1 \text{ m/s}^2}$ toward the centre
✍ IN-TEXT PRACTICE 5.2 — NCERT classic (Cyclist on a circular track)
Q. A cyclist rides at 27 km/h on a circular track of radius 80 m and applies brakes,
reducing speed at 0.5 m/s². Find the magnitude and direction of his net acceleration at that instant.
When speed varies along a circular path of radius $r$:
Radial / Centripetal Acceleration ($a_c$): Responsible for changing direction:
$$a_c = \frac{v^2}{r} = \omega^2 r \quad (\text{directed toward centre})$$
Tangential Acceleration ($a_t$): Responsible for changing speed magnitude:
$$a_t = \frac{dv}{dt} = r\alpha = r\frac{d\omega}{dt} \quad (\text{directed along tangent})$$
Net Total Acceleration ($a_{\text{net}}$): Since $\vec{a}_c \perp \vec{a}_t$:
$$a_{\text{net}} = \sqrt{a_c^2 + a_t^2} = \sqrt{\left(\frac{v^2}{r}\right)^2 + \left(\frac{dv}{dt}\right)^2}$$
$$\tan\phi = \frac{a_t}{a_c} \quad (\phi = \text{angle of } \vec{a}_{\text{net}} \text{ with radius vector})$$
Conical Pendulum
A bob of mass $m$ suspended by a string of length $L$ revolving in a horizontal circle of radius $r = L\sin\theta$ with constant angular speed $\omega$:
∴ $R = \tfrac12 g\,t_1t_2$ ✓ (a very frequent proof question)
✍ TYPE 6 — JEE Main · River Crossing at Minimum Drift ($v_r > v_b$ case)
Q. A swimmer can swim at 3 m/s in still water; the river flows at 5 m/s. At what angle
should he swim to minimise drift? (Note $v_r > v_b$ — zero drift impossible.)
Result (derived via calculus/geometry): aim upstream at angle θ
with the perpendicular where $\sin\theta = \dfrac{v_b}{v_r} = \dfrac{3}{5} \Rightarrow \theta =
37°$ — i.e., swim at 90° + 37° = 127° from the river-flow direction.
Contrast with Section 6.1: when $v_b > v_r$ the zero-drift condition is
$\sin\theta = v_r/v_b$; when $v_b < v_r$ the minimum-drift condition flips to $\sin\theta =
v_b/v_r$. Both are asked in JEE.
✍ TYPE 7 — NEET · Ratio of Centripetal Accelerations
Q. Two particles move on circles of radii in ratio 1 : 2. Compare their centripetal
accelerations when (a) their speeds are equal, (b) their time periods are equal.
(b) Same T: $a_c = 4\pi^2 r/T^2 \propto r \Rightarrow a_1 : a_2
= \mathbf{1 : 2}$
Moral: pick the formula whose "same" quantity is given — the most common
UCM MCQ pattern.
✍ TYPE 8 — HC Verma Level · Projectile Hitting a Wall (Coordinate Attack)
Q. A ball is thrown at 20 m/s at 53° from the foot of a vertical wall 12 m away (sin 53° =
0.8, cos 53° = 0.6, g = 10 m/s²). At what height does it hit the wall?
Method moral: any "hits a wall / clears a fence / lands on a roof" question =
find t from the horizontal equation, substitute in the vertical equation.
7. Common Mistakes & Misconceptions — Final Checklist
Night Before Exam
Read this the night before the exam:
In 2-D projectile motion, $u\cos\theta$ never changes; only $v_y$ evolves. Time $t$ is the only parameter linking horizontal and vertical equations.
At the apex (highest point) of a projectile: $v \ne 0$ (it equals $u\cos\theta$, the minimum speed of flight); only $v_y = 0$. The acceleration is always $g$ vertically downward.
Complementary launch angles ($\,\theta\,$ and $\,90^\circ - \theta\,$) yield identical range $R$, but different maximum heights ($H_1, H_2$) and times of flight ($T_1, T_2$). Note that $R = 4\sqrt{H_1 H_2} = \tfrac12 g T_1 T_2$.
Time of flight of a horizontal projectile from height $h$ is $\sqrt{2h/g}$ — completely independent of the initial horizontal speed $u$.
A body dropped or released from a moving carrier (plane, balloon, drone) inherits the carrier's velocity vector at that exact instant.
River crossing time depends solely on the velocity component perpendicular to the river bank ($t = d / (v_b\cos\theta)$).
Zero drift across a river is physically possible only when swimmer speed exceeds river speed ($v_b > v_r$). If $v_r > v_b$, only minimum drift is possible.
Uniform Circular Motion (UCM) has constant speed, but it is continuously accelerated motion because velocity direction changes every instant. Work done by centripetal force is strictly zero.
In Non-Uniform Circular Motion (NUCM), total acceleration has two perpendicular components: radial $a_c = v^2/r$ and tangential $a_t = dv/dt$. Thus $a_{\text{net}} = \sqrt{a_c^2 + a_t^2}$.
Landing speed from a tower of height $h$ is always $\sqrt{u^2 + 2gh}$ regardless of the angle of projection (by conservation of mechanical energy).
VARDAAN COMET