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Class 11 Physics Notes

Motion in a Plane Class 11 Physics Notes

Complete, exam-ready notes on motion in a plane: vectors and their operations, projectile motion, and uniform circular motion — written for CBSE, JEE and NEET revision.

Class11SubjectPhysicsCoversCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is projectile motion in one line?

Projectile motion is the two-dimensional motion of an object under constant gravity, where horizontal motion is uniform and vertical motion is uniformly accelerated, giving a parabolic trajectory.

Vectors and Their Operations

  • A vector has magnitude and direction; scalars have only magnitude.
  • Dot product:
  • A⃗⋅B⃗=∣A⃗∣∣B⃗∣cos⁡θ\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos\theta
  • (scalar).
  • Cross product:
  • ∣A⃗×B⃗∣=∣A⃗∣∣B⃗∣sin⁡θ|\vec{A}\times\vec{B}| = |\vec{A}||\vec{B}|\sin\theta
  • (vector perpendicular to both).
  • Component method: add x- and y-components separately.
  • Unit vector has magnitude 1 and gives direction.
A⃗⋅B⃗=∣A∣∣B∣cos⁡θ,∣A⃗×B⃗∣=∣A∣∣B∣sin⁡θ\vec{A}\cdot\vec{B} = |A||B|\cos\theta,\quad |\vec{A}\times\vec{B}| = |A||B|\sin\theta
Dot and cross products

Projectile Motion

Projectile equations

x=ucos⁡θ t,y=usin⁡θ t−12gt2x = u\cos\theta\,t,\quad y = u\sin\theta\,t - \frac{1}{2}gt^2

For a projectile launched at angle θ with speed u: horizontal motion (no acceleration) is uniform; vertical motion is uniformly accelerated under gravity g.

  • Time of flight:
  • T=2usin⁡θgT = \frac{2u\sin\theta}{g}
  • .
  • Maximum height:
  • H=u2sin⁡2θ2gH = \frac{u^2\sin^2\theta}{2g}
  • .
  • Range:
  • R=u2sin⁡2θgR = \frac{u^2\sin 2\theta}{g}
  • .
  • Range is maximum (R_max = u²/g) at 45°.
  • Two angles (θ and 90°−θ) give the same range for the same speed.
H=u2sin⁡2θ2g,R=u2sin⁡2θg,T=2usin⁡θgH = \frac{u^2\sin^2\theta}{2g},\quad R = \frac{u^2\sin 2\theta}{g},\quad T = \frac{2u\sin\theta}{g}
Projectile key results

Velocity at any instant

At time t the instantaneous velocity is v=(ucos⁡θ)2+(usin⁡θ−gt)2v = \sqrt{(u\cos\theta)^2 + (u\sin\theta - gt)^2}, and it is always tangential to the trajectory.

Uniform Circular Motion

  • Uniform circular motion has constant speed but changing velocity (direction changes), so it is accelerated.
  • Centripetal acceleration:
  • ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r
  • , always toward the centre.
  • Angular velocity:
  • ω=2πT=2πf\omega = \frac{2\pi}{T} = 2\pi f
  • .
  • Tangential acceleration is zero in uniform circular motion.
ac=v2r=ω2r,ω=2πTa_c = \frac{v^2}{r} = \omega^2 r,\quad \omega = \frac{2\pi}{T}
Uniform circular motion

Relative Velocity

Relative velocity

v⃗AB=v⃗A−v⃗B\vec{v}_{AB} = \vec{v}_A - \vec{v}_B

The relative velocity of A with respect to B is the difference of their velocities. It is used in river-boat and wind-aircraft problems.

Solved Examples

Example: A projectile is launched at 30∘30^\circ with u=20 m/su = 20\,\text{m/s}. Find the maximum height. Take g=10 m/s2g = 10\,\text{m/s}^2.

Solution: H=u2sin⁡2θ2g=400×(1/4)20=5 mH = \frac{u^2\sin^2\theta}{2g} = \frac{400\times(1/4)}{20} = 5\,\text{m}.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Max height

H=u2sin⁡2θ2gH = \frac{u^2\sin^2\theta}{2g}

Range

R=u2sin⁡2θgR = \frac{u^2\sin 2\theta}{g}

Time of flight

T=2usin⁡θgT = \frac{2u\sin\theta}{g}

Centripetal acceleration

ac=v2ra_c = \frac{v^2}{r}

Angular velocity

ω=2πT\omega = \frac{2\pi}{T}

Relative velocity

vAB=vA−vBv_{AB} = v_A - v_B

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Range maximum at 45° (R = u²/g).
  • Two angles θ and 90°−θ give the same range.
  • Uniform circular motion is accelerated (direction changes).
  • Centripetal acceleration always points to the centre.
  • Projectile path is parabolic.

FAQ

Common questions

Why is a projectile's path parabolic?

Horizontal motion is uniform (constant velocity) while vertical motion is uniformly accelerated (by g). Their combination gives x ∝ t and y ∝ t², which produces a parabola.

What is the angle for maximum range in projectile motion?

45°, giving R = u²/g. At any other angle the range is smaller because sin(2θ) is maximised at 2θ = 90°.

Is uniform circular motion accelerated?

Yes — even with constant speed, the velocity changes because its direction changes continuously, so there is a centripetal acceleration toward the centre.

What is relative velocity?

The velocity of one object as seen from another: v_AB = v_A − v_B. It is the basis for river-boat and wind-aircraft velocity problems.

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All Class 11 NotesClass 11 Laws of Motion NotesClass 11 Units and Measurements NotesPhysics Tutor in Mohali

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