Ultimate Guide to Projectiles and Resisted Motion in HSC Mathematics Extension 2

Introduction

In HSC Mathematics Extension 1, you learned that a projectile fired into the air follows a perfect, symmetrical parabolic path. However, in HSC Mathematics Extension 2, we burst that bubble by introducing air resistance. Projectiles and Resisted Motion is the ultimate test of your mechanics knowledge, requiring you to resolve vectors in 2D space while simultaneously solving complex differential equations for both the horizontal and vertical components of motion.

Executive Summary

This comprehensive guide covers the core concepts of 2D resisted projectile motion:

• Trajectory Comparison: Distinguishing between an idealized unresisted parabola and the asymmetric, curtailed path of a resisted projectile.
• Establishing Equations of Motion: Setting up horizontal ($m\ddot{x} = -km\dot{x}$) and vertical ($m\ddot{y} = -mg - km\dot{y}$) differential equations for resistance proportional to speed.
• Integrating for Displacement: Solving these equations to find horizontal and vertical displacement as functions of time.
• Solving Problems: Using the derived $x(t)$ and $y(t)$ equations to find time of flight, range, and maximum height.

What is this about?

When you hit a golf ball, it doesn't travel in a perfect parabola. Air resistance pushes against it the entire time. The faster it goes, the harder the air pushes back. In Extension 2, we model this by assuming the resistive force is proportional to the projectile's velocity vector.

By splitting the problem into horizontal and vertical components, we can track the projectile's progress. You will find that the horizontal velocity decays exponentially, meaning the projectile essentially hits an invisible "wall" (a maximum possible range) if given enough time. The vertical motion behaves exactly like the Vertical Resisted Motion we studied previously.

Main Content

1. Shape of the Trajectory

An unresisted projectile has a constant horizontal velocity and a constant vertical acceleration ($-g$). The resulting path is a symmetrical parabola.

A resisted projectile experiences deceleration in both directions.

• Horizontal: The horizontal speed $\dot{x}$ decays exponentially.
• Vertical: The vertical speed $\dot{y}$ decays faster on the way up and reaches a terminal velocity on the way down.
• Trajectory Shape: The path is asymmetrical. The angle of descent is much steeper than the angle of projection, and the total range is significantly shorter than an unresisted projectile. As $t \to \infty$, the projectile approaches a vertical asymptote (it falls straight down).

2. Equations of Acceleration

Consider a particle of mass $m$ projected from the origin $(0,0)$ with initial velocity $V$ at an angle $\theta$ to the horizontal. It moves under gravity ($g$) and experiences air resistance of magnitude $kmv$ (where $v$ is the speed and $k$ is a constant).

We resolve the forces into horizontal ($x$) and vertical ($y$) components. Let upwards and rightwards be positive.

Horizontal Motion: The only force is the horizontal component of air resistance. $m\ddot{x} = -km\dot{x}$ $\ddot{x} = -k\dot{x}$

Vertical Motion: Gravity acts downwards, and the vertical component of air resistance acts opposite to the vertical velocity. $m\ddot{y} = -mg - km\dot{y}$ $\ddot{y} = -g - k\dot{y}$

Note: The syllabus specifically requires you to be able to handle resistance proportional to the first power of speed ($R \propto v$) for projectiles, because $R \propto v^2$ results in non-linear coupled differential equations that cannot be easily solved by hand.

3. Deriving Expressions for Displacement

You are expected to derive $x(t)$ and $y(t)$ from scratch in an exam.

Horizontal Displacement $x(t)$: Start with $\ddot{x} = -k\dot{x}$. $\frac{d\dot{x}}{dt} = -k\dot{x}$ $\int \frac{1}{\dot{x}} d\dot{x} = \int -k \, dt$ $\ln(\dot{x}) = -kt + C_1$

Initial conditions: At $t=0$, horizontal velocity $\dot{x} = V\cos\theta$. $C_1 = \ln(V\cos\theta)$ $\ln(\dot{x}) - \ln(V\cos\theta) = -kt$ $\dot{x} = V\cos\theta \, e^{-kt}$

Integrate again to find $x$: $x = \int V\cos\theta \, e^{-kt} \, dt$ $x = -\frac{V\cos\theta}{k} e^{-kt} + C_2$

Initial conditions: At $t=0, x=0$. $0 = -\frac{V\cos\theta}{k}(1) + C_2 \implies C_2 = \frac{V\cos\theta}{k}$ Final $x(t)$: $x(t) = \frac{V\cos\theta}{k} (1 - e^{-kt})$

Vertical Displacement $y(t)$: Start with $\ddot{y} = -(g + k\dot{y})$. $\frac{d\dot{y}}{dt} = -(g + k\dot{y})$ $\int \frac{1}{g + k\dot{y}} d\dot{y} = \int -1 \, dt$ $\frac{1}{k}\ln(g + k\dot{y}) = -t + C_3$

Initial conditions: At $t=0, \dot{y} = V\sin\theta$. $C_3 = \frac{1}{k}\ln(g + kV\sin\theta)$ Substitute and rearrange: $\ln\left(\frac{g + k\dot{y}}{g + kV\sin\theta}\right) = -kt$ $g + k\dot{y} = (g + kV\sin\theta)e^{-kt}$ $\dot{y} = \frac{g + kV\sin\theta}{k}e^{-kt} - \frac{g}{k}$

Integrate again to find $y$: $y = \int \left( \frac{g + kV\sin\theta}{k}e^{-kt} - \frac{g}{k} \right) dt$ $y = -\frac{g + kV\sin\theta}{k^2}e^{-kt} - \frac{gt}{k} + C_4$

Initial conditions: At $t=0, y=0$. $0 = -\frac{g + kV\sin\theta}{k^2}(1) - 0 + C_4 \implies C_4 = \frac{g + kV\sin\theta}{k^2}$ Final $y(t)$: $y(t) = \frac{g + kV\sin\theta}{k^2}(1 - e^{-kt}) - \frac{gt}{k}$

Simple Worked Example

Question: Using the equations derived above, find an expression for the maximum horizontal range of the projectile as time $t \to \infty$.

Solution: The horizontal displacement equation is: $x(t) = \frac{V\cos\theta}{k} (1 - e^{-kt})$

To find the theoretical maximum range, we take the limit as $t \to \infty$. Since $k > 0$, as $t \to \infty$, the term $e^{-kt} \to 0$.

Therefore: $\lim_{t \to \infty} x(t) = \frac{V\cos\theta}{k} (1 - 0) = \frac{V\cos\theta}{k}$

Answer: The projectile can never travel further than $\frac{V\cos\theta}{k}$ metres horizontally, regardless of how high it is fired or how long it falls. This is a defining characteristic of resisted projectiles!

mini-FAQ page

Q: Do I need to memorize those massive $x(t)$ and $y(t)$ formulas? A: No! Just like with vertical resisted motion, you are expected to derive them from $m\ddot{x} = -km\dot{x}$ and $m\ddot{y} = -mg - km\dot{y}$. Memorizing them is dangerous because a question might change the coordinate system (e.g., firing from a cliff where $y(0) = H$).

Q: Will we ever be asked to solve projectiles where resistance is proportional to $v^2$? A: In HSC Extension 2, you will only be asked to solve 2D projectiles where $R \propto v$. If $R \propto v^2$, the horizontal and vertical components cannot be easily separated (because the velocity vector magnitude $|v| = \sqrt{\dot{x}^2 + \dot{y}^2}$ couples the equations together), which is beyond the scope of high school calculus.

Common mistakes to avoid

• Forgetting the constants of integration: This is the #1 mistake. When integrating to find $x$ and $y$, students often assume $C=0$ because the particle starts at the origin. As shown in the derivations above, $C$ is rarely zero in these exponential equations. Always substitute $t=0$ to find $C$ explicitly.
• Confusing $\dot{x}$ and $x$: Make sure your notation is spotless. It is very easy to drop a dot and accidentally integrate velocity as if it were displacement.

Practice on Vu's Maths Hub

Resisted projectiles represent the peak of algebraic and calculus integration in the HSC. Put your skills to the ultimate test:

• Practice full derivations for projectile motion in the HSC Mechanics Booklet.
• Solidify your exponential and logarithmic integration techniques with the HSC Integrals Booklet.
• Take on the toughest 4-unit exam-style mechanics questions in the HSC Last Resorts Booklet.

Further Readings

• Review the foundations of vertical resistance in our guide on Vertical Resisted Motion.
• Explore more HSC math resources and full worked solutions at Vu's Maths Hub.

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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!