Motion in a Plane – Comprehensive Notes on 2-D Motion and Projectile Motion

2-D Path Shapes from Parametric Equations

Core idea: A trajectory in a plane is the locus of (x, y) traced as time t varies for given x(t) and y(t).
1-D straight line paths arise when x and y share the same time dependence up to a scale.
- Example: x = a sin t, y = b sin t ⇒ y = (b/a) x (straight line through the origin).
2-D curved paths arise when the coordinates do not share a single proportionality.
- Elliptical path: x = a sin t, y = b cos t ⇒ x^2/a^2 + y^2/b^2 = 1.
- Circle path (special case of ellipse): x = R cos t, y = R sin t ⇒ x^2 + y^2 = R^2.
- Parabolic/path that looks like a parabola: e.g., r = 2 t i + 4 t^2 j ⇒ x = 2t, y = 4t^2 ⇒ y = x^2.
Practice examples from the transcript:
- Ellipse: x = a sin t, y = b cos t ⇒ x^2/a^2 + y^2/b^2 = 1.
- Circle: x^2 + y^2 = R^2 (parametrization x = R cos t, y = R sin t).
- Parabola: x = α t, y = β t^2 ⇒ y = (β/α^2) x^2.

Method: If velocity is V = ⟨Vx, Vy⟩, then dx/dt = Vx and dy/dt = Vy.
The trajectory satisfies dy/dx = (dy/dt)/(dx/dt) = Vy/Vx.
Example 1: V = ⟨2, x⟩
- dx/dt = 2, dy/dt = x ⇒ dy/dx = x/2
- Integrate: dy = (x/2) dx ⇒ y = x^2/4 + C
- Through origin: C = 0 → y = x^2/4.
Example 2: V = ⟨y, x⟩
- dx/dt = y, dy/dt = x ⇒ dy/dx = x/y
- Solve: y dy = x dx ⇒ (1/2) y^2 = (1/2) x^2 + C → y^2 − x^2 = C'
- Through origin: C' = 0 → y^2 = x^2 → y = ± x.

Decompose initial velocity: ux = u cos θ, uy = u sin θ.
Equations of motion (under constant gravity g downward):
- x(t) = u cos θ · t
- y(t) = u sin θ · t − (1/2) g t^2
Trajectory equation (eliminate t):
- t = x/(u cos θ) ⇒ y = x tan θ − (g x^2)/(2 u^2 cos^2 θ)
Key quantities (for motion from ground to ground, initial height 0):
- Time of flight: Tf = (2 u sin θ)/g
- Horizontal range: R = (u^2 sin 2θ)/g
- Maximum height: H = (u^2 sin^2 θ)/(2 g)
Example with consistent numbers from the transcript:
- Let θ = 45°, choose u such that ux = 40 m/s and uy = 40 m/s. This occurs if u = 40√2 m/s and θ = 45°.
- With g = 10 m/s^2:
- Tf = 8 s
- R = 320 m
- H = 80 m
- Initial speed |v0| = u = 40√2 m/s
- Velocity over time: V(t) = ⟨u cos θ, u sin θ − g t⟩ = ⟨40, 40 − 10t⟩
- At t = 4 s: V = ⟨40, −40⟩
- Direction of velocity at t = 1 s: tan φ = (uy − g t)/ux = (40 − 10)/40 = 0.75 → φ ≈ 36.87°
- Time when velocity is perpendicular to the initial velocity: V · V0 = 0 → t = Tf = 8 s (perpendicular at the end of flight here)
Trajectory in x-y form (explicit): y(x) = x tan θ − (g x^2)/(2 u^2 cos^2 θ)
Note on air resistance: If drag acts only along x or has a simple model, the y-motion still governed by gravity; realistic air resistance modifies the trajectory, but the standard result is for vacuum-like conditions.

Problem 1: Horizontal range equals maximum height
- R = (u^2 sin 2θ)/g, H = (u^2 sin^2 θ)/(2 g)
- Set R = H: sin 2θ = (1/2) sin^2 θ
- Using sin 2θ = 2 sin θ cos θ: 2 sin θ cos θ = (1/2) sin^2 θ
- For sin θ ≠ 0: tan θ = 4 → θ = tan^{-1}(4)
Problem 2: Range is 4√3 times maximum height
- R/H = [u^2 sin 2θ / g] / [u^2 sin^2 θ /(2 g)] = [2 sin 2θ] / [sin^2 θ] = 4 cot θ
- Solve 4 cot θ = 4√3 → cot θ = √3 → θ = 30°
Problem 3: Bullet with speed u at angle θ; compute H, T, R
- H = (u^2 sin^2 θ)/(2 g)
- T = (2 u sin θ)/g
- R = (u^2 sin 2θ)/g
- Example (from transcript): u = 280 m/s, θ = 30°, g ≈ 9.8 m/s^2
- sin θ = 1/2, sin 2θ = sin 60° = √3/2
- H ≈ (280^2 × 0.25)/(2 × 9.8) ≈ 1000 m
- T ≈ (2 × 280 × 0.5)/9.8 ≈ 28.6 s
- R ≈ (280^2 × √3/2)/9.8 ≈ large value (compute with exact numbers if needed)
- Important: Here results depend on g; the transcript’s numerical options may assume a rounding or different g value.

From x(t) and y(t) you can usually eliminate t to get y as a function of x.
General form (no air resistance):
- If x = u cos θ t, then t = x/(u cos θ)
- Substitute into y: y = x tan θ − (g x^2)/(2 u^2 cos^2 θ)
This is the standard projectile trajectory equation in vacuum.

Example A: Position vector r = 2 t i + 4 t^2 j
- x = 2t, y = 4t^2 → t = x/2 → y = 4 (x/2)^2 = x^2 → Trajectory: y = x^2 (parabola).
Example B: Velocity field V = ⟨2, x⟩
- dx/dt = 2, dy/dt = x → dy/dx = x/2 → y = x^2/4 + C (through origin → y = x^2/4).
Example C: Velocity field V = ⟨y, x⟩
- dx/dt = y, dy/dt = x → dy/dx = x/y → y dy = x dx → (1/2) y^2 = (1/2) x^2 + C → y^2 − x^2 = C'
- Through origin → y^2 = x^2 → y = ± x.
Example D: Elliptical path from parametric form
- x = a sin t, y = b cos t → x^2/a^2 + y^2/b^2 = 1.
Example E: Circular path
- x^2 + y^2 = R^2.

Trajectory shapes from simple parametric forms:
- Straight line: x = a sin t, y = b sin t ⇒ y = (b/a) x.
- Ellipse: x = a sin t, y = b cos t ⇒ x^2/a^2 + y^2/b^2 = 1.
- Circle: x = R cos t, y = R sin t ⇒ x^2 + y^2 = R^2.
- Parabola (from linear x with quadratic y): y ∝ x^2.
Velocity to trajectory conversion: dy/dx = Vy/Vx.
Projectile motion (no air resistance):
- x(t) = u cos θ · t, y(t) = u sin θ · t − (1/2) g t^2
- Trajectory: y = x tan θ − (g x^2)/(2 u^2 cos^2 θ)
- Key quantities: Tf = 2 u sin θ / g, R = u^2 sin 2θ / g, H = u^2 sin^2 θ /(2 g)
Special relationships:
- If horizontal range equals maximum height: θ = tan^{-1}(4).
- If R = 4√3 × H: θ = 30°.
- Relationship between max height and range (for same launch conditions): H = (R/4) tan θ.
Velocity and motion notes from transcript examples:
- For u = 40√2 m/s and θ = 45°, g = 10 m/s^2:
- Tf = 8 s, R = 320 m, H = 80 m
- Initial speed |v0| = 40√2 m/s
- V(t) = ⟨40, 40 − 10 t⟩
- At t = 4 s: V = ⟨40, −40⟩
- Perpendicular condition V · V0 = 0 occurs at t = 8 s in this setup.

// Notation recap for exam-style problems

u: magnitude of initial velocity, θ: launch angle, g: acceleration due to gravity
ux = u cos θ, uy = u sin θ
R = (u^2 sin 2θ)/g, H = (u^2 sin^2 θ)/(2g), Tf = (2 u sin θ)/g
Trajectory equation: y = x tan θ − (g x^2)/(2 u^2 cos^2 θ)
Time to reach a given vx or vy can be found by solving x = u cos θ t and y = u sin θ t − (1/2) g t^2 for t and substituting back as needed