Chapter 3 Readings: 3.5 Free Fall

Fundamentals of Free Fall and Gravity

Free fall describes the motion of an object falling strictly under the influence of a gravitational field, such as near the surface of Earth or other planetary celestial bodies. In idealized free fall, motion is assumed to occur along a straight vertical line perpendicular to the planetary surface, making it a one-dimensional kinematic process.

An everyday application of free-fall motion is estimating the depth of a vertical mine shaft by dropping a rock into it and measuring the elapsed time until the sound of impact is heard. Crucially, free fall is not limited to objects moving downward from a higher altitude to a lower altitude. An object thrown vertically upward is equally in a state of free fall during both its ascent and its descent, provided gravity is the only force acting upon it.

Mass Independence and Historical Context

When air resistance and friction are negligible, all objects in a given location fall toward the center of Earth with the exact same constant acceleration, regardless of their mass. Historically, prior to experiments by Galileo Galilei ($1564\text{--}1642$), it was widely believed that heavier objects fall with greater acceleration than lighter ones. Galileo demonstrated that in the absence of air resistance, light and heavy objects dropped simultaneously from the same height reach the ground at the exact same time.

This fundamental principle of gravity was demonstrated on the Moon in $1971$ by astronaut David R. Scott. In the lunar vacuum, where there is no atmospheric drag and acceleration due to gravity is only $1.67\,m/s^2$, Scott dropped a hammer and a feather simultaneously, and both hit the lunar surface at the exact same instant.

Real-World Drag and Resistance

In presence of air, aerodynamic drag and friction modify motion:

• Air resistance acts in opposition to the direction of motion through the atmosphere, causing lighter or less dense objects to fall slower than heavier or denser objects of similar size (e.g., a tennis ball reaches the ground after a baseball dropped simultaneously from a sufficient height).

• Contact friction opposes motion between solid or fluid interfaces, such as clothes rubbing against a laundry chute or a stone moving through a pool of water.

In ideal physics models, free fall strictly assumes the total absence of air resistance and friction.

Acceleration Due to Gravity ($g$)

The constant acceleration experienced by free-falling bodies is named the acceleration due to gravity, denoted by the symbol $g$.

• Standard and average values: Average magnitude at Earth's surface is $g = 9.81\,m/s^2$ (or $32.2\,ft/s^2$). In standard calculations with two significant figures, $g = 9.8\,m/s^2$ is used.

• Geographic variations: The magnitude of $g$ actually varies between $9.78\,m/s^2$ and $9.83\,m/s^2$ depending on latitude, altitude, local topography, and underlying geological formations.

• Direction: The direction of acceleration due to gravity is directed downward toward Earth's center. This downward gravitational direction strictly defines what is termed the vertical direction.

Kinematic Equations for One-Dimensional Free Fall

Vertical motion under constant gravitational acceleration is analyzed using one-dimensional kinematics. Vertical displacement is represented by the symbol $y$.

Boundary Conditions for Free Fall

1. Dropped Objects: An object released from rest has an initial vertical velocity $v_0 = 0\,m/s$.

2. Thrown Objects: An object thrown vertically possesses an initial vertical speed $v_0$ identical to its speed at the instant of release.

3. Start and End of Free Fall: An object enters free fall immediately after losing physical contact with whatever held or threw it. Free fall ends the instant the object touches the ground or another surface, at which point acceleration is no longer $g$.

Standard Equations (Upward Positive Coordinate System)

When defining the upward direction as positive and downward as negative, acceleration is $a = -g = -9.8\,m/s^2$:

$v = v_0 - gt$

$y = y_0 + v_0 t - \frac{1}{2}gt^2$

$v^2 = v_0^2 - 2g(y - y_0)$

If the coordinate system is inverted such that downward is positive, then acceleration is $a = g = 9.8\,m/s^2$.

Problem-Solving Strategy for Free Fall

1. Establish Coordinate Signs: Choose the sign convention for gravitational acceleration. Setting the upward direction as positive implies $a = -g$. Setting the downward direction as positive implies $a = g$.

2. Draw a Diagram: Sketch the physical setup showing trajectory, initial position, and final position.

3. Identify Variables: Record all known quantities ($y_0$, $y$, $v_0$, $v$, $t$, $g$) and identify the unknown variables.

4. Select Kinematic Equations: Match knowns and target unknowns to the appropriate kinematic equation ($v = v_0 - gt$, $y = y_0 + v_0 t - \frac{1}{2}gt^2$, or $v^2 = v_0^2 - 2g(y - y_0)$).

Worked Examples and Quantitative Applications

Example 3.14: Free Fall of a Ball Thrown Downward

A ball is thrown downward from the top of a $98.0\,m$ tall building with an initial velocity of $4.9\,m/s$.

Problem Formulation

• Origin $y_0 = 0\,m$ at the roof top.

• Upward direction positive, downward negative.

• Given: $y_0 = 0\,m$, $v_0 = -4.9\,m/s$, $y = -98.0\,m$, $g = 9.8\,m/s^2$.

Part (a): Elapsed Time to Ground

Using $y = y_0 + v_0 t - \frac{1}{2}gt^2$:

$-98.0\,m = 0 - (4.9\,m/s)t - \frac{1}{2}(9.8\,m/s^2)t^2$

$-98.0 = -4.9t - 4.9t^2$

$4.9t^2 + 4.9t - 98.0 = 0$

Dividing by $4.9$ yields the normalized quadratic polynomial:

$t^2 + t - 20 = 0$

Factoring $(t + 5.0)(t - 4.0) = 0$ gives roots:

$t = -5.0\,s \quad \text{and} \quad t = 4.0\,s$

• Physical Interpretation: $t = 0\,s$ is the time of release. The positive root $t = 4.0\,s$ is the physically valid duration for the ball to reach the ground. The negative root $t = -5.0\,s$ represents the hypothetical time a ball launched upward from ground level would have taken to reach the roof moving downward at $4.9\,m/s$.

Part (b): Impact Velocity

Using $v = v_0 - gt$ with $t = 4.0\,s$:

$v = -4.9\,m/s - (9.8\,m/s^2)(4.0\,s) = -44.1\,m/s$

• Significance: The velocity right before impact is $-44.1\,m/s$. Upon striking the ground, non-gravitational interaction forces decelerate the ball to zero velocity over a non-zero time interval.

Example 3.15: Vertical Motion of a Baseball

A batter hits a baseball straight upward from home plate. It is caught by the catcher $5.0\,s$ later at the same elevation.

Coordinate Setup

• Upward positive, origin $y_0 = 0\,m$ at home plate.

• $y = 0\,m$ at $t = 5.0\,s$.

Part (a): Initial Velocity ($v_0$)

$y = y_0 + v_0 t - \frac{1}{2}gt^2$

$0 = 0 + v_0(5.0\,s) - \frac{1}{2}(9.8\,m/s^2)(5.0\,s)^2$

$5.0 v_0 = 122.5 \implies v_0 = 24.5\,m/s$

Part (b): Maximum Height ($y_{max}$)

At peak height, instantaneous velocity $v = 0\,m/s$:

$v^2 = v_0^2 - 2g(y - y_0)$

$0 = (24.5\,m/s)^2 - 2(9.8\,m/s^2)(y - 0)$

$19.6 y = 600.25 \implies y = 30.6\,m$

Part (c): Time to Maximum Height

$v = v_0 - gt$

$0 = 24.5\,m/s - (9.8\,m/s^2)t \implies t = 2.5\,s$

• Symmetry Note: Ascent time ($2.5\,s$) equals descent time ($5.0\,s - 2.5\,s = 2.5\,s$).

Part (d): Acceleration at Maximum Height

At the apex, instantaneous velocity is zero ($v = 0\,m/s$), but acceleration remains constant at $9.8\,m/s^2$ downward ($a = -9.8\,m/s^2$).

Part (e): Catch Velocity

$v = v_0 - gt = 24.5\,m/s - (9.8\,m/s^2)(5.0\,s) = -24.5\,m/s$

• Significance: In free fall without air resistance, an object returns to its initial elevation with a speed equal to its launch speed, but opposite in direction.

Example 3.16: Rocket Booster Motion

A rocket blasts off vertically. At altitude $5.0\,km$ and upward velocity $200.0\,m/s$, it jettisons its booster.

Coordinate Setup

• Origin $y_0 = 0\,m$ selected at booster release point ($5.0\,km$ above ground).

• Upward direction positive, $v_0 = 200.0\,m/s$.

Part (a): Maximum Height above Release Point

At maximum height relative to release point, $v = 0\,m/s$:

$v^2 = v_0^2 - 2g(y - y_0)$

$0 = (200.0\,m/s)^2 - 2(9.8\,m/s^2)y$

$y = \frac{v_0^2}{2g} = \frac{(2.0 \times 10^2\,m/s)^2}{2(9.8\,m/s^2)} = 2040.8\,m \approx 2.04\,km$

• Total altitude above Earth's surface = $5.0\,km + 2.04\,km \approx 7.0\,km$.

Part (b): Booster Velocity at Altitude $6.0\,km$

An altitude of $6.0\,km$ above Earth corresponds to $y = 1.0 \times 10^3\,m$ relative to the release point origin:

$v^2 = v_0^2 - 2g(y - y_0)$

$v^2 = (200.0\,m/s)^2 - 2(9.8\,m/s^2)(1.0 \times 10^3\,m)$

$v^2 = 40000 - 19600 = 20400\,m^2/s^2$

$v = \pm 142.8\,m/s$

• Significance: $v = +142.8\,m/s$ represents upward velocity during ascent at $6.0\,km$ altitude; $v = -142.8\,m/s$ represents downward velocity at the same altitude during descent.