Motion in a Plane

1. 2-D or 3-D Motion

• We divide 2-D or 3-D motion into 2 or 3 mutually perpendicular directions (axes) and solve the motion on each axis separately.
• Motion only along x-axis → 1-D ; in x–y plane → 2-D ; in x–y–z space → 3-D.

2. Kinematics

In 3D (vector form) :

Along different axes :

(1-D : a = dvdt = vdvdx ,  v = dxdt)

Axis	Velocity	Acceleration
x	vx = dx/dt	ax = dvx/dt = vx dvx/dx
y	vy = dy/dt	ay = dvy/dt = vy dvy/dy
z	vz = dz/dt	az = dvz/dt = vz dvz/dz

3. Uniform Acceleration Motion

(a→ = constant vector)

i.e. v² = u² + 2 a→·s→

4. Rectangular Components (2D)

• Any vector (v, a, F) can be split into two perpendicular components.

• Component along the side of the angle → cosθ ; the other one → sinθ.

5. Addition of Rectangular Components

6. Projectile – Assumptions

• Ignore air resistance.
• g is uniform (constant, downward).
• Only force on the body = mg (downward) ⇒ acceleration = g downward.

7. Ground to Ground Projectile

• A particle (mass m) is projected from the ground with speed u at an angle θ with the horizontal. Only force = mg ⇒ acceleration = g downward.

Symbol	Meaning
T	time of flight (A → B → C)
H	maxm height of ball
R	range = distance AC
t₁	time of ascent (A → B)
t₂	time of descent (B → C)

• At highest point B, vertical velocity = 0 (only u cosθ remains).

8. Motion along x & y

x-axis :

• u_x = u cosθ ,  a_x = 0 ⇒ velocity constant in x-direction

y-axis :

• u_y = u sinθ ,  a_y = −g (constant) ⇒ v_y changes

9. Velocity at any time

• v_x = u cosθ = constant ;  v_y = u sinθ − gt → variable
• Net speed and angle φ change with time :

10. Time of Flight (T)

• Best tool : s = ut + ½at² in y-direction (A to C, y = 0) :

0 = u sinθ · T + ½(−g)T²

11. Time of Ascent (t₁)

• v = u + at in y-direction (A to B, v_y = 0) :

0 = u sinθ + (−g) t₁

12. Maximum Height (H)

• v² = u² + 2as in y-direction (A to B) :

0² = (u sinθ)² + 2(−g)(H)

13. Range (R)

• x-direction : s = ut (a = 0)

R = (u cosθ) T = (u cosθ)2u sinθg

14. Fixed Speed u, Angle θ Variable

• T and H are maximum for θ = 90° (sinθ = 1) → thrown vertically up.
• Range is maximum when sin2θ = 1 ⇒ θ = 45° :

• Range is same for angles θ and (90° − θ) :

R′ = u² sin2(90° − θ)g = u² sin(180° − 2θ)g

    = u² sin2θg = R

15° & 75° → same R ; 30° & 60° → same R ; 45° → R_max

15. Equation of Trajectory

• Equation of path of the particle (eliminate t) :

x = (u cosθ) t ⇒ t = xu cosθ

y = (u sinθ)xu cosθ − ½ gx²u² cos²θ

• Path of a projectile is a parabola (downward opening).

16. Formula Summary (Ground to Ground)

Quantity	Formula
Time of flight	T = 2u sinθ / g = 2uy/g
Max. height	H = u² sin²θ / 2g = uy²/2g
Range	R = u² sin2θ / g = 2uxuy/g
Ascent / descent	t₁ = t₂ = T/2
Trajectory	y = x tanθ − gx²(1 + tan²θ)/2u²

• Let u fixed, θ varies : θ = 90° → T, H max^m ; θ = 45° → R max^m ; θ and 90° − θ → R same.
• Particle lands with the same speed u at the same angle θ (below horizontal).
• At the same height h, speed is the same while going up and coming down.

17. Tower to Ground Projectile

• Ball thrown straight up with u sinθ and ball thrown at angle θ with u :
• Both motions are different, but same in y-direction (difference only in x-direction) ⇒ same time to reach the ground.
• Use y-direction : −H = (u sinθ)T − ½gT²

18. Horizontal Projectile

• Body thrown horizontally with speed u from height H (u_y = 0).

• y-direction : s = ut + ½at² ⇒ −H = 0·T + ½(−g)T²

• Time of flight is the same as free fall from height H (u = 0) – it doesn't depend on u.
• Vertical speed at ground = √(2gH) in both cases.

19. Motion Graphs (Solved)

• u = 40 m/s, a = −10 m/s²
• T = 2u_yg = 8 s ,  t₁ = t₂ = 4 s
• H = u_y²2g = (40)²20 = 80 m

(1) v–t graph :  v = 40 − 10t

Straight line : slope = −10, intercept = +40

(2) H–t graph :  H = 40t − 5t²

Downward parabola : t = 0 → H = 0 ; t = 4 → H = 80 m ; t = 8 → H = 0

(3) v–H graph :  v² = 1600 − 20H

Parabola (opening towards −H) : v = 40 → H = 0 ; v = 0 → H = 80 ; v = −40 → H = 0

20. Projectile on Inclined Plane

• Incline at angle α with horizontal ; ball projected at angle θ with the incline (φ = θ + α with the horizontal).

Method 1 : Intersect trajectory with the incline line

x tanφ − gx²2u²(1 + tan²φ) = x tanα

⇒ tanφ − tanα = gx2u²(1 + tan²φ)

• Solve for x, then y = x tanα ; time from x = (u cosφ)t ; range along incline R = √(x² + y²).

Method 2 : Change the axes

• Take x-axis along the incline and y-axis ⊥ to the incline. Resolve g :

Axis	Initial velocity	Acceleration
x (along incline)	u cosθ	∓ g sinα
y (⊥ incline)	u sinθ	− g cosα

21. Up the Incline

• u_x = u cosθ , a_x = −g sinα ;  u_y = u sinθ , a_y = −g cosα (both constant)
• Time of flight : y-direction, s = 0 :
  0 = u sinθ T + ½(−g cosα)T²

• Range (x-direction) :
  R = u cosθ · T − ½ g sinα · T²

22. Down the Incline

• Projected from the top towards the bottom ; θ with the incline.
• Now a_x = +g sinα (along the motion down the incline), a_y = −g cosα.

• Range : R = u cosθ · T + ½ g sinα · T²

23. Equation of Envelope

• A cannon (or fountain) fires with fixed speed u at all angles.
• The envelope is the boundary – no shell / water drop can go outside it, for any angle.

Derivation :

• Trajectory : y = x tanθ − gx²2u²(1 + tan²θ)

gx²2u²tan²θ − x tanθ + y + gx²2u² = 0

• If the particle can pass through (x, y), this equation gives real roots of tanθ ⇒ D ≥ 0 :

x² − 4gx²2u²(y + gx²2u²) ≥ 0

⇒ 1 − 2gyu² − g²x²u⁴ ≥ 0  ⇒  y ≤ u²2g − gx²2u²

• It is a downward parabola.
• x = 0 ⇒ y = u²2g = H_max  ;  y = 0 ⇒ x = ± u²g = R_max
• Point (x, y) inside the envelope → reachable by two angles ; on it → one angle ; outside → not reachable.

24. Parabolic Curves

• y = ax² + bx + c : a +ve → opens upward ; a −ve → opens downward.
• x = ay² + by + c : a +ve → opens towards +x ; a −ve → towards −x.
• e.g. H = 40t − 5t² (a −ve) → downward parabola ; v² = 1600 − 20H → opens towards −H.

25. Quick Formula Revision

Case	Formula
Time of flight	2u sinθ / g
Max. height	u² sin²θ / 2g
Range	u² sin2θ / g ; Rmax = u²/g
Trajectory	y = x tanθ − gx²/(2u² cos²θ)
Horizontal projectile	T = √(2H/g), R = u√(2H/g)
Incline (time)	2u sinθ / (g cosα)
Envelope	y = u²/2g − gx²/2u²