Free Fall Motion – Study Notes

Objectives

  • Describe the concept of free-fall motion.
  • Appreciate the pervasive role of gravity in everyday life and cultivate curiosity about natural free-fall phenomena.
  • Apply the standard kinematic equations to solve quantitative free-fall problems.

Drill: “Gravity Alone?” (Concept Check)

  • 1. Rock dropped from a cliff → YES (solely gravity).
  • 2. Bird flying in the sky → NO (wing‐generated lift & drag).
  • 3. Feather in a vacuum → YES (air resistance absent).
  • 4. Book falling from a table → YES (air resistance negligible if ignored).
  • 5. Paper airplane → NO (aerodynamic lift & drag).

Core Concept: Free-Fall Motion

  • An object is in free fall when gravity is the only force acting on it.
    • Air drag, wind, or any propulsion must be ignored/absent.
  • On Earth the (near-surface) gravitational acceleration is
    g = -9.8\;\text{m/s}^2
    (negative sign ⇒ downward direction in a standard upward-positive coordinate system).
  • All objects, regardless of mass or composition, share the same acceleration in free fall (Galileo’s insight; validated by Apollo 15 hammer–feather experiment on the Moon).

Everyday & Philosophical Relevance

  • Explains why raindrops, falling apples, thrown balls, stones from cliffs, etc. accelerate downward.
  • Underpins safety equipment design (airbags, helmets) and architectural codes (e.g., calculating impact speeds from rooftops).
  • Sparks philosophical reflection on universality of natural laws and equality of inertial & gravitational mass (foundation of Einstein’s equivalence principle).

Kinematic Framework (for 1-D Vertical Motion)

  • We reuse the constant-acceleration formulas, substituting a with g.
  • Sign convention: “Up” is positive; hence a=g=-9.8\;\text{m/s}^2.

Distance Fallen (Time Known)

  • Formula: d = v_i t + \frac{1}{2} a t^2
    • Useful when release velocity v_i and elapsed time t are given.
    • If object is dropped (from rest): v_i = 0 → d = \tfrac{1}{2} a t^2.

Distance (Time Unknown)

  • When final velocity is known instead of time: vf^2 = vi^2 + 2 a d
    • Rearrange to solve for d: d = \dfrac{vf^2 - vi^2}{2a}.

Time of Fall (Velocity Route)

  • If initial & final velocities are known:
    vf = vi + a t → t = \dfrac{vf - vi}{a}.

Distance in Terms of Average Velocity

  • Combining definitions: d = \frac{vi + vf}{2} \, t
    • Highlights that displacement equals average velocity × time under constant acceleration.

Worked Examples

Example 1 – Vase Thrown Upward

  • Given: vi = 26.2\,\text{m/s},\; vf = 0,\; a = -9.8\,\text{m/s}^2.
  • Target: Maximum height.
  • Use vf^2 = vi^2 + 2 a d ⇒
    0 = (26.2)^2 + 2(-9.8)d ⇒
    d = \frac{- (26.2)^2}{2(-9.8)} = 35.02\,\text{m} \approx 35\,\text{m}.
  • Physical insight: upward motion slows until velocity zero, then reverses.

Example 2 – Shingles Dropped from Roof

  • Given: d = -8.52\,\text{m},\; v_i = 0,\; a = -9.8\,\text{m/s}^2.
  • Solve d = \frac{1}{2} a t^2 ⇒
    t = \sqrt{\frac{2|d|}{|a|}} = \sqrt{\frac{2\times 8.52}{9.8}} = 1.32\,\text{s}.
  • Takeaway: drop time depends only on height (for constant g).

Example 3 – Downward Launch, Find v_f

  • Given: v_i = -2\,\text{m/s},\; d = -45\,\text{m},\; a = -9.8\,\text{m/s}^2.
  • Apply vf^2 = vi^2 + 2 a d:
    vf^2 = (-2)^2 + 2(-9.8)(-45) = 4 + 882 = 886, vf = -\sqrt{886} \approx -29.7\,\text{m/s}.
  • Negative sign ⇒ downward direction.

Example 4 – Drop from 20 m

  • Step A (Time): t = \sqrt{\frac{2\times 20}{9.8}} = 2.02\,\text{s}.
  • Step B (Final Velocity): v_f = 0 + (-9.8)(2.02) = -19.8\,\text{m/s}.
  • Interpretation: impact speed ≈ 20\,\text{m/s} (~72 km/h).

Practice Problems (Activity #2)

  • 1. Ball thrown straight up with v_i = 18\,\text{m/s}. Question: Total time in the air?
    • Hint: Time up = Time down; top occurs when v=0.
  • 2. Ball dropped from 20\,\text{m} building. Question: Impact speed?
    • Use Example 4 or v_f^2 = 2 g d.
  • 3. Ball thrown upward vi = 18\,\text{m/s}. a) Highest height? b) Time to reach that height? (use vf=0 at peak).

Quick Reference: Key Equations

  • Displacement: d = v_i t + \frac{1}{2} a t^2.
  • Final velocity: vf = vi + a t.
  • Velocity–displacement link: vf^2 = vi^2 + 2 a d.
  • Average velocity form: d = \frac{vi + vf}{2} t.

Practical Safety Implications

  • Engineering of guardrails, balcony heights, and sports safety nets depends on maximum potential energy m g h & impact velocity \sqrt{2 g h}.
  • Understanding fall times crucial for timing parachute deployment, drone drops, and fair-ground rides.

Conceptual Connections

  • Builds on introductory constant-acceleration kinematics (lecture 1).
  • Prepares ground for later topics: projectile motion (adds horizontal component), Newton’s laws (forces), and energy conservation (potential ↔ kinetic).

Summary

  • Free fall = motion under gravity alone.
  • All objects share g, hence motion is mass-independent.
  • Four kinematic formulas suffice to analyze time, displacement, and velocity.
  • Sign convention clarity prevents common sign-errors.
  • Real-world design, safety, and scientific inquiry hinge on these principles.