🎯 Projectile Range — derive once, use forever
Posted: Oct 19, 2025 • Tags: kinematics, vectors
A quick derivation for the range on level ground. Know your assumptions, then reuse this elegantly.
Given: launch speed v at angle θ; flat ground; ignore air resistance.
We want: horizontal range R.
Break velocity: vₓ = v cosθ, vᵧ = v sinθ.
Time to return to y=0: t = 2 vᵧ / g.
Range: R = vₓ · t = (v cosθ)(2 v sinθ / g) = v² sin(2θ) / g.
Show solution & checks
- Units: v²/g → (m²/s²)/(m/s²) = m ✅
- Max range at θ=45°: sin(90°)=1 → Rmax=v²/g ✅
- Small-angle limit: sin(2θ)≈2θ → R≈(2v²θ)/g ✅
🧰 Free‑Body Diagram Mastery — the 5‑line method
Posted: Oct 19, 2025 • Tags: forces, newton-2
Stop guessing forces. Choose a system, choose axes, and let Newton’s 2nd law do the talking.
1) Choose the system and draw only forces on it.
2) Choose axes that simplify (e.g., along the incline).
3) Write ΣF = m a along each axis.
4) Connect constraints (no slip, rope taut → a's equal).
5) Solve symbolically first; numbers second.
Example: block on incline (no friction)
Along-slope: ΣF = m a → m g sinθ - N·0 = m a → a = g sinθ
Normal: ΣF = 0 → N = m g cosθ
🔋 Energy is a Story (not just numbers)
Posted: Oct 19, 2025 • Tags: energy, pedagogy
Teaching energy like bookkeeping misses the poetry. Tell the story: “Where did it come from? Where did it go?”
Framework:
• Define your system clearly (include/exclude Earth, springs, fields).
• Identify transfers (work, heat) vs transformations (KE ↔ PE ↔ internal).
• Write ΔE_system = W_on + Q, then track terms visually with arrows.
Mini-Example
Drop a mass from height h (no air):
m g h (PE) → ½ m v² (KE). Internal, Q, W_by hand = 0.