Ultimate Guide to Vertical Resisted Motion in HSC Mathematics Extension 2

Introduction

Throwing a ball straight up into the air seems like a simple physics problem, but when you factor in air resistance, the mathematics becomes significantly more complex. In HSC Mathematics Extension 2, Vertical Resisted Motion requires you to combine the constant downward pull of gravity with a variable resistive force that changes direction depending on whether the object is moving up or down. Mastering this topic requires sharp integration skills and a deep understanding of Newtonian physics.

Executive Summary

This comprehensive guide covers the core concepts of vertical resisted motion:

Equations of Motion: Setting up $m\ddot{x} = \pm mg \pm kv^n$ based on the direction of travel (upwards vs. downwards).
Deriving Velocity and Displacement: Using calculus ( $a = \frac{dv}{dt}$ and $a = v\frac{dv}{dx}$ ) to find $v(t)$ , $v(x)$ , and $x(t)$ .
Terminal Velocity: Understanding and calculating the constant speed an object reaches when resistance perfectly balances gravity.
Maximum Height and Flight Time: Finding the peak of the trajectory ( $v=0$ ) and the total time taken to return to the ground.

What is this about?

When an object moves vertically, gravity is always pulling it down. Air resistance, however, always opposes the direction of motion. If you throw a ball up, both gravity and air resistance pull it down, resulting in rapid deceleration. When the ball falls back down, gravity pulls it down, but air resistance pushes up. Because the forces change depending on the direction of travel, you cannot use a single equation for the entire flight. You must split the problem into an "upward journey" and a "downward journey."

Main Content

1. Setting up the Equations of Motion

Always start by defining your positive direction. For vertical motion, defining upwards as positive is usually standard, but you can define downwards as positive if the object is only falling.

Case A: Moving Upwards (Let Upwards be Positive)

Velocity $v$ is positive.
Gravity $mg$ acts downwards (negative).
Resistance $kv^n$ opposes motion, acting downwards (negative).
Equation: $m\ddot{x} = -mg - kv^n$

Case B: Moving Downwards (Let Downwards be Positive)

Velocity $v$ is positive (since we defined downwards as positive).
Gravity $mg$ acts downwards (positive).
Resistance $kv^n$ opposes motion, acting upwards (negative).
Equation: $m\ddot{x} = mg - kv^n$

Crucial Note: The sign of the resistance term depends entirely on your chosen coordinate system and the direction of the particle's velocity.

2. Terminal Velocity

When an object falls (Case B), gravity accelerates it downwards, which increases its speed. As speed increases, the resistance ( $kv^n$ ) increases. Eventually, the upward resistance perfectly balances the downward gravitational force. When $mg = kv^n$ , the net force is zero, meaning acceleration $\ddot{x} = 0$ .

The velocity at which this occurs is called the terminal velocity, $V_T$ . $0 = mg - kV_T^n \implies V_T = \left(\frac{mg}{k}\right)^{\frac{1}{n}}$

If $R \propto v$ ( $n=1$ ), then $V_T = \frac{mg}{k}$ . If $R \propto v^2$ ( $n=2$ ), then $V_T = \sqrt{\frac{mg}{k}}$ .

Tip: Many HSC questions will ask you to rewrite the equation of motion in terms of $V_T$ instead of $k$ and $m$ . For example, if $k = \frac{mg}{V_T^2}$ , then $m\ddot{x} = mg - \frac{mg}{V_T^2}v^2 = mg(1 - \frac{v^2}{V_T^2})$ .

3. Deriving Expressions for Velocity and Displacement

Once you have your equation of motion, you integrate it exactly as you did for rectilinear motion, but with the added $mg$ term.

To find velocity as a function of time ( $v(t)$ ), use $a = \frac{dv}{dt}$ . $m\frac{dv}{dt} = mg - kv^2$ $\frac{dv}{mg - kv^2} = \frac{1}{m} dt$ (This often requires integrating using inverse hyperbolic functions or partial fractions).

To find velocity as a function of displacement ( $v(x)$ ), use $a = v\frac{dv}{dx}$ . $m v \frac{dv}{dx} = -mg - kv^2$ (for upward motion) $\frac{v}{mg + kv^2} dv = -\frac{1}{m} dx$ (This usually integrates to a natural logarithm, $\ln$ ).

4. Maximum Height and Flight Time

To find the maximum height, analyze the upward journey. The particle reaches maximum height when its velocity is zero ( $v=0$ ). Use your $v(x)$ equation, set $v=0$ , and solve for $x$ .

To find the time to reach maximum height, use your $v(t)$ equation for the upward journey, set $v=0$ , and solve for $t$ .

To find the time to return to the ground, you must set up a new equation of motion for the downward journey, derive the $x(t)$ equation for the downward journey, set $x$ to the maximum height (or $0$ depending on your origin), and solve for $t$ . The total flight time is the sum of the upward time and downward time.

Simple Worked Example

Question: A particle of mass $m$ is dropped from rest (so it only travels downwards). It experiences air resistance of magnitude $mkv^2$ , where $v$ is its speed and $k$ is a positive constant. Let downwards be the positive direction. (a) Find its terminal velocity $V_T$ . (b) Find an expression for velocity $v$ in terms of displacement $x$ .

Solution: (a) Find Terminal Velocity Let downwards be positive. Gravity acts downwards (positive). Resistance acts upwards (negative). $F_{\text{net}} = mg - mkv^2$ $ma = mg - mkv^2$ $a = g - kv^2$

Terminal velocity occurs when acceleration is zero ( $a=0$ ). $0 = g - k V_T^2$ $k V_T^2 = g$ $V_T = \sqrt{\frac{g}{k}}$

(b) Find $v$ in terms of $x$ Start with $a = g - kv^2$ . We want $v$ and $x$ , so we use $a = v\frac{dv}{dx}$ . $v\frac{dv}{dx} = g - kv^2$

Separate variables: $\frac{v}{g - kv^2} dv = dx$

Integrate both sides: $\int \frac{v}{g - kv^2} dv = \int dx$ Notice the derivative of the denominator is $-2kv$ . We adjust the numerator: $-\frac{1}{2k} \int \frac{-2kv}{g - kv^2} dv = \int dx$ $-\frac{1}{2k} \ln(g - kv^2) = x + C$

Find $C$ using initial conditions. The particle is dropped from rest, so at $x=0, v=0$ . $-\frac{1}{2k} \ln(g - 0) = 0 + C \implies C = -\frac{1}{2k} \ln(g)$

Substitute $C$ back: $-\frac{1}{2k} \ln(g - kv^2) = x - \frac{1}{2k} \ln(g)$ $-\frac{1}{2k} [ \ln(g - kv^2) - \ln(g) ] = x$ $\ln\left(\frac{g - kv^2}{g}\right) = -2kx$ $\frac{g - kv^2}{g} = e^{-2kx}$ $g - kv^2 = g e^{-2kx}$ $kv^2 = g(1 - e^{-2kx})$ $v^2 = \frac{g}{k}(1 - e^{-2kx})$

Since $V_T^2 = \frac{g}{k}$ , we can elegantly rewrite this as: $v^2 = V_T^2(1 - e^{-2kx})$

mini-FAQ page

Q: Why do we have to split the motion into "up" and "down"? A: Because the formula for the net force changes. Going up, gravity and resistance are in the same direction. Going down, they are in opposite directions. A single differential equation cannot capture a sudden change in force direction.

Q: Is terminal velocity always reached? A: Mathematically, $v \to V_T$ as $t \to \infty$ . It is an asymptote. In practice, an object will get extremely close to terminal velocity if it falls for long enough, but theoretically, it only ever approaches it.

Common mistakes to avoid

Integrating $\frac{1}{a^2 + x^2}$ incorrectly: In upward motion with $R \propto v^2$ , you'll often integrate $\frac{1}{g + kv^2}$ . This requires an inverse tangent ( $\tan^{-1}$ ), not a logarithm! Only use $\ln$ when you have a $v$ in the numerator (like $\frac{v}{g+kv^2}$ ).
Forgetting to redefine the origin for the downward journey: If you find the maximum height $H$ , it is often easiest to place a new origin at the top of the trajectory and define downwards as positive for the return trip, rather than keeping the origin on the ground.

Practice on Vu's Maths Hub

Vertical resisted motion requires you to synthesize everything you know about physics, calculus, and algebra.

Practice deriving $v(t)$ and $x(t)$ equations step-by-step in the HSC Mechanics Booklet.
Review inverse trigonometric integration with the HSC Integrals Booklet.
Tackle the most brutal 4-unit terminal velocity proofs in the HSC Last Resorts Booklet.

Connect with me

Want to master Mechanics and lock in top marks for HSC Mathematics Extension 2? Visit Vu's Maths Hub for in-depth booklets, detailed video explanations, and expert advice to help you ace your exams!