Unlike motion in a straight line, real-world objects — such as a kicked football, a thrown javelin, a satellite orbiting Earth, or a car turning on a curved track — move in two dimensions. Motion in a Plane forms Chapter 4 of the CBSE Class 11 Physics syllabus and is a foundational scoring topic for JEE Main, JEE Advanced, and NEET.
This comprehensive guide covers vector algebra (addition, resolution, dot and cross products), oblique projectile motion derivations (trajectory, time of flight, max height, range), horizontal projectile motion, uniform circular motion, centripetal acceleration, relative velocity in 2D (rain-man and river-swimmer problems), and five step-by-step solved entrance exam numericals.
- 1. Vector Algebra Foundations: Addition, Resolution & Products
- 2. Oblique Projectile Motion: Assumptions & Derivations
- 3. Projectile Motion Master Formulas Summary Table
- 4. Horizontal Projectile Motion (Projection from a Height)
- 5. Uniform Circular Motion (UCM) & Centripetal Acceleration
- 6. Relative Velocity in Two Dimensions (Rain-Man & River-Swimmer)
- 7. Solved Entrance Exam Numerical Problems (JEE / NEET)
- 8. Frequently Asked Questions (FAQ)
1. Vector Algebra Foundations: Addition, Resolution & Products
A vector quantity has both magnitude and direction, and obeys vector addition laws.
• Magnitude of Resultant (R):
R = √(A² + B² + 2AB cosθ)
• Direction (α with A):
tan α = (B sinθ) / (A + B cosθ)
A · B = AB cosθ (e.g. Work W = F · s)
i · i = j · j = k · k = 1 | i · j = 0
• Cross Product (Vector):
|A × B| = AB sinθ n_hat (e.g. Torque τ = r × F)
i × i = 0 | i × j = k
2. Oblique Projectile Motion: Assumptions & Derivations
A Projectile is an object thrown into space with an initial velocity under the influence of gravity alone. Assumed: no air resistance, flat Earth, uniform acceleration g downward.
Suppose a particle is projected with velocity u at an angle θ with the horizontal:
- Horizontal Component: Initial u_x = u cosθ, Acceleration a_x = 0 (Horizontal velocity remains constant!).
- Vertical Component: Initial u_y = u sinθ, Acceleration a_y = -g (Decelerates upward, accelerates downward).
1. Horizontal displacement: x = (u cosθ) t ⇒ t = x / (u cosθ).
2. Vertical displacement: y = (u sinθ) t - ½g t².
3. Substituting t: y = (u sinθ)(x / (u cosθ)) - ½g (x / (u cosθ))².
y = x tanθ - (g x²) / (2 u² cos²θ) or y = x tanθ (1 - x / R)
This is the equation of a PARABOLA! Thus, the path of a projectile is parabolic.
3. Projectile Motion Master Formulas Summary Table
| Parameter | Mathematical Formula | Key Concept & Maximum Condition |
|---|---|---|
| Time of Flight (T) | T = (2 u sinθ) / g | Total time spent in air. Time of ascent t_a = t_d = (u sinθ) / g. |
| Maximum Height (H) | H = (u² sin²θ) / (2g) | Reached when vertical velocity v_y = 0. Maximum at θ = 90° (H_max = u² / (2g)). |
| Horizontal Range (R) | R = (u² sin 2θ) / g | Total horizontal distance. Maximum Range at θ = 45° (R_max = u² / g). |
| Complementary Angles Rule | R(θ) = R(90° - θ) | Two projection angles θ and 90° - θ (e.g. 30° and 60°) produce equal ranges! |
| Velocity at any time t | v = √(v_x² + v_y²) = √[(u cosθ)² + (u sinθ - gt)²] | Direction: tan φ = v_y / v_x. At highest point, v_min = u cosθ. |
4. Horizontal Projectile Motion (Projection from a Height)
When a body is projected horizontally with velocity u from height h (u_x = u, u_y = 0, a_y = -g):
• Horizontal Range: R = x = u · t = u √(2h / g).
• Equation of Trajectory: y = (g x²) / (2 u²) (Parabolic path).
• Velocity at impact: v = √(u² + 2gh) at angle tan φ = √(2gh) / u.
5. Uniform Circular Motion (UCM) & Centripetal Acceleration
When a body moves along a circular path of radius r with constant speed v, its direction changes continuously, creating a constant magnitude acceleration directed toward the center called Centripetal Acceleration (a_c).
ω = dθ / dt = 2π / T = 2π f (rad/s)
• Linear vs Angular Velocity:
v = ω r
a_c = v² / r = ω² r = v ω
• Centripetal Force:
F_c = m a_c = (m v²) / r = m ω² r
6. Relative Velocity in Two Dimensions (Rain-Man & River-Swimmer)
Angle to hold umbrella: tan θ = v_m / v_r with vertical.
2. River-Swimmer Problem: Swimmer speed in still water v_s, river velocity v_r, river width d:
• To cross in SHORTEST TIME: Swim directly perpendicular (θ = 90°). t_min = d / v_s. Drift x = v_r · t_min.
• To cross along SHORTEST PATH (zero drift): Swim upstream at angle sin θ = v_r / v_s (requires v_s > v_r). Time t = d / √(v_s² - v_r²).
7. Solved Entrance Exam Numerical Problems (JEE / NEET)
(a) H = (u² sin²θ) / (2g) = (28)² × (0.5)² / (2 × 9.8) = (784 × 0.25) / 19.6 = 196 / 19.6 = 10 m.
(b) T = (2 u sinθ) / g = (2 × 28 × 0.5) / 9.8 = 28 / 9.8 = 2.86 s.
(c) R = (u² sin 60°) / g = (784 × 0.866) / 9.8 = 80 × 0.866 = 69.28 m.
(a) t = √(2h / g) = √[(2 × 490) / 9.8] = √(980 / 9.8) = √(100) = 10 s.
(b) Vertical velocity at impact v_y = g t = 9.8 × 10 = 98 m/s.
Resultant speed v = √(v_x² + v_y²) = √(15² + 98²) = √(225 + 9604) = √(9829) = 99.14 m/s.
For angle (90° - θ): R2 = [u² sin 2(90° - θ)] / g = [u² sin(180° - 2θ)] / g = (u² sin 2θ) / g = R1. (Hence Proved!).
Maximum height ratio: H1 = (u² sin²θ) / (2g), H2 = [u² sin²(90° - θ)] / (2g) = (u² cos²θ) / (2g).
Ratio H1 / H2 = (sin²θ) / (cos²θ) = tan²θ.
Frequency f = 7 / 100 = 0.07 rev/s.
Angular speed ω = 2π f = 2 × (22/7) × 0.07 = 2 × 22 × 0.01 = 0.44 rad/s.
Linear speed v = ω r = 0.44 × 0.12 = 0.053 m/s = 5.3 cm/s.
Relative velocity of rain w.r.t woman: v_rw = v_r - v_w.
tan θ = v_w / v_r = 12 / 35 ≈ 0.343 ⇒ θ = tan¯¹(0.343) ≈ 19°.
Explore Related CBSE Physics & Lab Apparatus Guides
8. Frequently Asked Questions (FAQ)
The path (trajectory) of a projectile launched at an angle with horizontal is a Parabola, represented by equation y = x tanθ - (g x²) / (2 u² cos²θ).
The horizontal range is maximum at a projection angle of 45°, where R_max = u² / g.
Centripetal acceleration is the inward acceleration experienced by a body moving in a circular path, required to continuously change velocity direction. Formula: ac = v² / r = ω² r.
In Uniform Circular Motion, the magnitude of velocity (speed) remains constant, but its direction changes continuously at every point along the circular path, creating acceleration.
The Dot Product A · B = AB cosθ results in a scalar (e.g. Work). The Cross Product |A × B| = AB sinθ results in a vector perpendicular to both vectors (e.g. Torque).
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