Modelling with quadratics

In a nutshell

Mathematical models aim to represent real-life scenarios as closely as possible. Given the difficulty to achieve total accuracy, the models are often only valid under certain circumstances or only for certain input values. Quadratics offer an example of an equation that can model practical contexts, such as projectile motion.

Notation

You will be used to working with the variables $x$ and $y$ when working with quadratics. For example, the following is a quadratic equation and can be sketched:

$y=-x^2+8x$

Maths; Quadratics; KS5 Year 12; Modelling with quadratics

But when modelling real-life scenarios, often other letters are used - ones that make more sense in the context of the scenario. For example, instead of $x$ , often $t$ is used to represent time. Instead of $y$ , sometimes $d$ is used for displacement, or $h$ for height. Hence, the following is an equivalent quadratic:

$h=-t^2+8t$

When used for modelling, $h$ can be given as a function in terms of $t$ :

$h(t)=-t^2+8t$

Moreover, if representing real-life, some constraints can be given. Here, if $t$ represents time, then $t\geq0$ since the model is meaningless for negative time.

Modelling a projectile

Quadratics are useful for modelling the height (or displacement from the start point) against the time taken of a projectile (anything thrown).

Example 1

A projectile is launched into the air from atop a hill and lands on the ground below the hill. It's height from the ground is modelled by

$h(t)=-t^2+8t+9$ 

where $t\geq0$ and represents time from when the projectile is launched. Interpret the meaning of the constant term $9$ in this equation and then sketch the journey of the projectile.

Given that the function $h$ represents height from the ground, when $t=0$ (i.e. just before the projectile is launched), the projectile is at a height

$h(0)=9$ 

So the $9$ represents the initial height - in other words, it's the height of the hill.

To sketch the projectile's journey, plot the equation $h(t)=-t^2+8t+9$ . The coordinate grid will have a vertical $h$ -axis and a horizontal $t$ -axis. Sketching can be done in the same way you know how to sketch a quadratic, but ensure that $t\geq0$ and that the curve does not go beneath the $t$ -axis, since you are told that the projectile lands on the ground. In other words, this means that $h\geq0$ .

Maths; Quadratics; KS5 Year 12; Modelling with quadratics

Example 2

A projectile's journey is modelled by the equation

$h(t)=3t-t^2$

where $h$ represents the height in metres of the projectile at the time $t$ after launch.

i. From what height is the projectile launched?

ii. How long does it take for the projectile to return to the ground?

iii. What is the maximum height of the projectile?

i. When $t=0$ , the projectile is launched. Inserting this into the equation gives $h(0)=0$ .

The launch height is $\underline{0m}$ .

ii. You are looking for the times when $h=0$ . Insert this into the equation and solve for $t$ :

$\begin {aligned}0&=3t-t^2 \\ 0&=t(3-t)\\t&=0,t=3\end {aligned}$

You know that the launch is at $t=0$ , so the projectile must return to the ground at the other solution, when $t=3$ .

The projectile lands after $\underline{3s}$ .

iii. To find the maximum height, you need to find the turning point (also known as the vertex) of the quadratic curve. This can be done by completing the square:

$\begin {aligned}h(t)&=-t^2+3t\\h(t)&=-(t^2-3t)\\h(t)&=-((t-1.5)^2-(-1.5)^2)\\h(t)&=-(t-1.5)^2+2.25\end {aligned}$ 

The turning point has coordinates $(1.5,2.25)$ .

The maximum height is $\underline{2.25m}$ .