Falling Objects with Drag — Falling Objects with Linear...

SECTION 1

Historical Context & Motivation

For centuries, thinkers assumed that heavier objects fall faster than lighter ones. Aristotle taught this idea around 350 BCE, and it went largely unchallenged for almost two thousand years. It was not until the Renaissance that scientists began to question this assumption by performing actual experiments and developing the mathematics to describe motion more precisely.

When Galileo Galilei showed that all objects in a vacuum accelerate at the same rate, he set the stage for a deeper question: why don't objects in the real world behave exactly that way? The answer lies in air resistance, or drag — a force that opposes motion through a fluid. Modeling drag mathematically requires a differential equation, making this one of the most natural first applications of first-order ODEs.

~350 BCE

Aristotle's Claim

Aristotle asserts that heavier objects fall proportionally faster, an idea that dominates Western thought for nearly two millennia.

1589

Galileo's Falling-Body Experiments

Galileo demonstrates that, ignoring air resistance, all objects accelerate at the same rate regardless of mass. His inclined-plane experiments formalize the concept of uniform acceleration.

1687

Newton's Principia

Isaac Newton publishes his laws of motion and proposes that resistive forces in fluids can be proportional to the velocity of the moving object, laying the groundwork for linear drag models.

1851

Stokes' Law

George Gabriel Stokes derives a formula for the drag force on a small sphere moving slowly through a viscous fluid, confirming that drag is proportional to velocity at low speeds.

The central question this lesson addresses is: how do we write and solve a differential equation that predicts the velocity of a falling object when air resistance is proportional to velocity? Understanding this model connects Newtonian physics to the powerful techniques of first-order ODEs.

SECTION 2

Core Principles & Definitions

Before diving into the math, you need to understand a few foundational ideas. In free fall without air resistance, the only force on an object is gravity. The moment we introduce a fluid — like air or water — a new resistive force appears that depends on how fast the object is moving.

Gravity (Weight)

The downward force mg acting on an object of mass m, where g ≈ 9.8 m/s². This force is constant near Earth's surface.

Linear Drag Force

A resistive force proportional to velocity: F_drag = −bv, where b is the drag coefficient and v is the speed. The negative sign means it opposes motion.

Newton's Second Law

The net force on an object equals its mass times its acceleration: F_net = ma. This is the bridge that turns a physics setup into a differential equation, since acceleration is dv/dt.

Terminal Velocity

The maximum constant speed reached when the drag force exactly balances gravity. At this point acceleration equals zero and dv/dt = 0. For linear drag, v_T = mg/b.

First-Order ODE

An equation involving a function and its first derivative. The falling-object equation m(dv/dt) = mg − bv is a first-order linear ODE in the unknown velocity v(t).

✦ KEY TAKEAWAY

Think of a skydiver jumping from a plane. At first, gravity dominates and the skydiver accelerates rapidly. As speed builds, air resistance grows until it matches gravity — like pressing harder on the brake of a car rolling downhill. Once the two forces balance, the skydiver stops accelerating and coasts at a constant terminal velocity. The differential equation we build in this lesson predicts exactly how velocity changes from the moment of release to the moment terminal velocity is reached.

SECTION 3

Free-Body Diagram & Force Balance

The best way to set up the differential equation is to start with a free-body diagram. The diagram below shows a ball falling downward with two forces acting on it: the weight mg pulling it down and the linear drag force bv pushing it up (opposing the motion). On the right, the diagram illustrates how these forces change as the object accelerates from rest toward terminal velocity.

Left: The free-body diagram shows the constant gravitational force mg (red, downward) and the velocity-dependent drag force bv (cyan, upward). Right: As time increases, drag grows while weight stays constant. When the two curves meet, net force is zero and terminal velocity is reached.

Notice a crucial feature on the right-hand graph: the red dashed line representing mg is perfectly flat because gravity doesn't depend on speed. The cyan curve for bv starts at zero (because the object starts from rest) and rises as the object speeds up. Where the two meet, the net force is zero, acceleration vanishes, and velocity becomes constant. This equilibrium speed is what we call the terminal velocity.

SECTION 4

Mathematical Framework

Let's convert the physics into a differential equation. We define the positive direction as downward (the direction of fall). The two forces on the object are: gravity mg (positive, downward) and drag −bv (negative, upward). Applying Newton's second law, F_net = ma, gives us the governing equation.

NEWTON'S SECOND LAW WITH LINEAR DRAG

m (dv/dt) = mg − bv

m = mass (kg), g = gravitational acceleration (≈ 9.8 m/s²), b = drag coefficient (kg/s), v = velocity (m/s), t = time (s).

Dividing both sides by m and rearranging puts this into standard first-order linear form:

STANDARD LINEAR FORM

dv/dt + (b/m)v = g

This is a first-order linear ODE in v(t). The coefficient b/m is constant, making this equation solvable with an integrating factor or separation of variables.

To solve by separation of variables, rewrite as dv/(g − (b/m)v) = dt. Integrating both sides (using a u-substitution on the left) and applying the initial condition v(0) = v₀ gives the general solution.

GENERAL VELOCITY SOLUTION

v(t) = (mg/b) + (v₀ − mg/b) e^(−bt/m)

If the object starts from rest (v₀ = 0), this simplifies to v(t) = (mg/b)(1 − e^−bt/m). Here mg/b is the terminal velocity v_T.

TERMINAL VELOCITY

v_T = mg / b

Set dv/dt = 0 in the original equation: 0 = mg − bv_T, so v_T = mg/b. Heavier objects or smaller drag coefficients lead to higher terminal velocities.

SECTION 5

Velocity vs. Time Graph & Time Constant

The solution v(t) = v_T(1 − e^−bt/m) describes an exponential approach to terminal velocity. The ratio τ = m/b is called the time constant. After one time constant, the velocity has reached about 63% of v_T; after three time constants, it has reached roughly 95%. The graph below makes this progression clear.

The cyan curve shows velocity rising from zero and asymptotically approaching the terminal velocity v_T (yellow dashed line). Purple dots mark key milestones: 63.2% at t = τ, 86.5% at t = 2τ, 95.0% at t = 3τ, and 98.2% at t = 4τ.

Velocity as a percentage of terminal velocity at key time intervals

Time (multiples of τ)	Percentage of v_T reached	Value of 1 − e^(−t/τ)
0	0%	0.000
1τ	63.2%	0.632
2τ	86.5%	0.865
3τ	95.0%	0.950
5τ	99.3%	0.993

The time constant τ = m/b controls how quickly the object approaches terminal velocity. A heavy object in a weakly resisting medium (large m, small b) has a large τ and takes a long time to reach terminal speed. A light object in thick fluid (small m, large b) reaches terminal velocity almost immediately.

SECTION 6

Worked Example

Let's walk through a complete problem. A small steel ball of mass 0.5 kg is dropped from rest through a viscous oil. The drag coefficient is b = 2.0 kg/s. Find the terminal velocity and the velocity after 0.5 seconds.

Steel Ball Falling Through Oil

Step 1 — Identify Given ValuesMass m = 0.5 kg, drag coefficient b = 2.0 kg/s, gravitational acceleration g = 9.8 m/s², initial velocity v₀ = 0 m/s (dropped from rest).

Step 2 — Calculate Terminal VelocityTerminal velocity is found by setting the net force to zero: v_T = mg/b = (0.5)(9.8) / 2.0 = 4.9 / 2.0.

v_T = 2.45 m/s

Step 3 — Calculate the Time ConstantThe time constant is τ = m/b = 0.5 / 2.0 = 0.25 s. This tells us that after 0.25 seconds, the ball has already reached about 63% of its terminal velocity.

τ = 0.25 s

Step 4 — Write the Velocity FunctionSince v₀ = 0, the solution simplifies to v(t) = v_T(1 − e^−t/τ) = 2.45(1 − e^−t/0.25).

Step 5 — Evaluate at t = 0.5 sSubstituting t = 0.5: v(0.5) = 2.45(1 − e^−0.5/0.25) = 2.45(1 − e⁻²) = 2.45(1 − 0.1353) = 2.45 × 0.8647.

v(0.5) ≈ 2.12 m/s (about 86.5% of terminal velocity)

Step 6 — Interpret the ResultNotice that t = 0.5 s is exactly 2τ. The table from Section 5 predicts 86.5% of terminal velocity at t = 2τ, which matches our calculation perfectly. The ball is nearly at top speed after only half a second because the drag coefficient is large relative to its mass.

SECTION 7

Strengths & Limitations of the Linear Drag Model

The linear drag model is elegant and solvable by hand, but it doesn't apply to every falling-object scenario. Understanding when it works well — and when it doesn't — is essential for choosing the right model in practice.

When to use (and not use) the linear drag model

Aspect	Strengths	Limitations
Solvability	Produces a clean, closed-form exponential solution — no numerical methods needed.	Only works when drag is proportional to v; quadratic drag (F ∝ v²) requires more advanced techniques.
Physical accuracy	Excellent for small objects at low speeds in viscous fluids (Stokes' law regime).	Poor for fast-moving objects in air (e.g., baseballs, skydivers) where drag scales with v².
Intuition building	Clearly introduces terminal velocity, time constants, and exponential approach — ideas that carry over to harder models.	May give students a false impression that all drag problems are equally straightforward.
Real-world examples	Pollen grains settling in air, tiny droplets in fog, microorganisms in water.	Not directly applicable to parachutes, cars, or most everyday objects without modification.

✦ KEY TAKEAWAY

Think of the linear drag model as training wheels for the broader topic of drag. Just as you learned to ride a bike with training wheels before tackling hills and turns, the linear model teaches you the essential concepts — terminal velocity, exponential approach, and the role of the drag coefficient — in a setting where the math is manageable. Once you've mastered these ideas, you'll be ready to handle the quadratic drag model and beyond.

SECTION 8

Connection to Advanced Theory: Quadratic Drag & Beyond

For most macroscopic objects falling through air at everyday speeds, the drag force is better described as proportional to the square of velocity: F_drag = −cv². This quadratic drag model leads to a different and harder differential equation. The table below compares the two models side by side.

Linear vs. quadratic drag comparison

Feature	Linear Drag (F = −bv)	Quadratic Drag (F = −cv²)
ODE	m(dv/dt) = mg − bv	m(dv/dt) = mg − cv²
Type	First-order linear	First-order nonlinear (separable)
Closed-form solution?	Yes — simple exponential	Yes — involves hyperbolic tangent (tanh)
Terminal velocity	v_T = mg/b	v_T = √(mg/c)
Best applies to	Tiny objects at low Re (Stokes flow)	Larger objects in air at moderate to high Re

The dimensionless quantity that determines which model to use is the Reynolds number (Re). When Re is small (roughly below 1), the linear model applies. When Re is large (above about 1000), quadratic drag dominates. In between, a more complex combination may be needed. As you continue studying differential equations, the quadratic drag model gives you practice with separable nonlinear ODEs and introduces the hyperbolic functions — a valuable extension of the ideas you've built here.

SECTION 9

Practice Problems

PROBLEM 1 — CONCEPTUAL

A falling object subject to linear drag eventually reaches terminal velocity. Explain, in terms of forces, why the object stops accelerating even though gravity never turns off.

PROBLEM 2 — BASIC CALCULATION

A 0.2 kg ball has a linear drag coefficient of b = 0.4 kg/s. Calculate the terminal velocity and the time constant τ.

PROBLEM 3 — INTERMEDIATE

Using the same ball from Problem 2 (m = 0.2 kg, b = 0.4 kg/s), find the velocity at t = 1.0 s assuming the ball is dropped from rest. Express your answer to two decimal places.

PROBLEM 4 — APPLIED

A small raindrop (mass 5.0 × 10⁻⁶ kg) falling through still air has a measured terminal velocity of 0.5 m/s. Determine the drag coefficient b and the time constant τ. Then estimate how long it takes the drop to reach 95% of its terminal velocity.

PROBLEM 5 — CRITICAL THINKING

Suppose an object is thrown downward with an initial speed v₀ that is greater than its terminal velocity v_T. Using the general solution v(t) = v_T + (v₀ − v_T)e^−bt/m, describe what happens to the velocity over time. Does the object speed up, slow down, or something else? Justify your answer using the equation.

SUMMARY

Lesson Summary

A falling object experiencing linear drag (F_drag = −bv) is governed by the first-order linear ODE m(dv/dt) = mg − bv. Solving this equation (via separation of variables or an integrating factor) yields v(t) = v_T(1 − e^(−t/τ)) for an object starting from rest, where v_T = mg/b is the terminal velocity and τ = m/b is the time constant.

The velocity rises exponentially toward v_T, reaching roughly 63% after one time constant and 95% after three time constants. This model works best for small objects at low Reynolds numbers (e.g., droplets, dust, microorganisms) and serves as a stepping stone to the more general quadratic drag model needed for larger everyday objects.