🥋 Kinematics in Two Dimensions – White to Black Belt Mastery

Quote

“Master one dimension, but true warriors move in two.”

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⚪ White Belt – Key Concept

In 2D motion, movement is split into horizontal and vertical components. These act independently, connected only by time. If you need a refresher on splitting vectors into x and y, review the Vectors Study Guide.

Core Principles

1. Break motion into x and y directions.
2. Horizontal velocity stays constant (no horizontal acceleration).
3. Vertical velocity changes due to gravity (g = 9.81 m/s² downward).
4. Use separate equations for x and y, then combine with vector math.

If you’re still solidifying the basics of motion, revisit Kinematics in One Dimension. For trigonometry skills needed here, check the Basic Math Study Guide.

Key Equations (SI Units)

• Horizontal:
  x = vₓ · t
• Vertical:
  y = vᵧ₀ · t − 0.5 g t²
• Velocity components:
  vₓ = v₀ cos θ
  vᵧ₀ = v₀ sin θ
• Time of flight (level ground):
  t = 2 vᵧ₀ / g
• Range:
  R = vₓ · t

Common Mistakes and Pitfalls

• Mixing up horizontal and vertical equations.
• Forgetting gravity only acts vertically.
• Using total velocity instead of components.

⚡ Sensei’s Shortcuts

• Always draw a diagram and mark vₓ, vᵧ₀, and θ.
• Time is the “bridge” between x and y motion.
• Treat horizontal and vertical like separate katas — then combine.

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✅ Worked Example – Step by Step (White Belt)

Problem: A ball is kicked with v₀ = 20.0 m/s at θ = 30°. Find the range.

Step 1. Formula
vₓ = v₀ cos θ
vᵧ₀ = v₀ sin θ
t = 2 vᵧ₀ / g
R = vₓ · t

Step 2. Substitution
vₓ = 20.0 cos 30° = 17.3 m/s
vᵧ₀ = 20.0 sin 30° = 10.0 m/s
t = 2(10.0) / 9.81 = 2.04 s
R = 17.3 × 2.04 = 35.3 m

Step 3. Final Answer
Range R = 35.3 m

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🏋️ Practice Drill (White Belt)

1. A projectile is launched at 15.0 m/s at 45°. Find the time of flight.
   Answer: 2.16 s
2. A rock is thrown horizontally at 12.0 m/s from a 20.0 m cliff. Find how far it travels before hitting the ground.
   Answer: 24.2 m

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🟡 Yellow Belt Extension – Deeper Skills

Problem: A projectile is launched at 25.0 m/s, θ = 40°. Find maximum height.

• Formula: H = (vᵧ₀²) / (2g)
• vᵧ₀ = 25.0 sin 40° = 16.1 m/s
• H = (16.1²) / (2 × 9.81) = 13.2 m

Final Answer: Maximum height = 13.2 m

Extra Practice

A ball is kicked at 18.0 m/s, θ = 60°. Find total flight time.
Answer: 3.18 s

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⚫ Black Belt Mastery – Exam Strategy and Challenge

Problem: A plane flies east at 50.0 m/s. A passenger throws a ball with speed 20.0 m/s at 45° above the horizontal (relative to the plane). Find the ground range from the release point.

Strategy Notes:

• Add plane velocity (50.0 m/s) to projectile’s horizontal component.
• Break problem into relative velocity + projectile motion.

Step 1. Velocity components
Relative vₓ = 20.0 cos 45° = 14.1 m/s
Total vₓ = 50.0 + 14.1 = 64.1 m/s
vᵧ₀ = 20.0 sin 45° = 14.1 m/s

Step 2. Time of flight
t = 2 vᵧ₀ / g = 2(14.1) / 9.81 = 2.87 s

Step 3. Range
R = vₓ · t = 64.1 × 2.87 = 184 m

Final Answer: Range = 184 m

If you need more review on the force principles behind motion, check Newton’s Second Law and Newton’s Third Law.

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🥋 Sensei’s Final Words

“In two dimensions, balance is everything. Split the motion, then bring it back together.”