## Putting It Together: Quadratic Functions

Now that you have seen how to work with quadratic functions in this module, let’s apply our skills to a problem involving projectile motion.

A trebuchet is a medieval weapon used to hurl large stones at enemy forts or castles in order to smash through the thick walls.  Basically the trebuchet works by putting the stone in a sling at the end of a long pole which is allowed to rotate around a fixed axis.  The shorter end of the pole is attached to a very heavy weight to counterbalance the projectile.  When the weight is released at the short end, the pole rotates, flinging the stone at high speed upward and toward the enemy.

Suppose a particular trebuchet can launch a 10 kilogram stone with an initial upward speed of 24.5 meters per second. Furthermore, suppose that the stone is 8 meters above the ground when it leaves the sling – that is, its initial height is 8 m. We can use what we know about quadratic functions to answer some interesting questions about the stone that is launched from the trebuchet:

1. What is the maximum height achieved by the stone during its flight to the enemy castle?
2. When does that maximum occur?
3. How long would it take before the stone comes crashing down to Earth (in case it missed the castle altogether)?

In order to answer these questions, we need to know about a particular quadratic function from physics called the ballistics equation:

$h(t)=-{\large\frac{g}{2}}t^2+v_0t+h_0$

Here, $h(t)$ stands for the height at time $t$, $v_0$ is the initial upward velocity (speed), $h_0$ is the initial height of the projectile, and $g$ is a constant called the acceleration due to gravity.  Near the surface of the Earth, $g\approx9.8$ meters per second squared, so $-{\Large\frac{g}{2}}\approx-4.9$. Now putting the given values into their proper places, we have:

$h(t)=-4.9t^2+24.5t+8$

This is a quadratic function with $a=-4.9$, $b=24.5$, $c=8$.  The graph is a parabola, and it will have a maximum because $a<0$.  The maximum function value in this case represents the maximum height of the stone.

Now we can answer the first two questions, What is the maximum height achieved by the stone during its flight to the enemy castle, and at what time does this occur?  To find the maximum height and time at which it occurs, we use the vertex formula which will give $(t_\text{max},h(t_\text{max}))$. the time at maximum point and height at the time of the maximum point.

$t_\text{max}=-{\large\frac{b}{2a}}=-{\large\frac{24.5}{2(-4.9)}}=2.5$

$h(t_\text{max})=h\left(-{\large\frac{b}{2a}}\right)=h(2.5)=-4.9(2.5)^2+24.5(2.5)+8\approx38.6$

Therefore, the maximum height will be 38.6 meters, which occurs 2.5 seconds after launch.  That’s a high-flying projectile!

Finally, to determine when the stone hits the ground, we just have to find the $x$-intercept of the function.  In other words, we have to solve $h(t)=0$.  That’s a job for the quadratic formula.

$t_\text{final}={\large\frac{-b\pm\sqrt{b^2-4ac}}{2a}}={\large\frac{-24.5\pm\sqrt{24.5^2-4(4.9)(8)}}{2a}}\approx-0.3,5.3$

As expected, the quadratic formula gives us two answers, but only the positive one makes sense in this context.  The stone lands on the ground about 5.3 seconds after it was launched.

Quadratic functions can be used to model the behavior of objects in free fall, amongst other things. We can use algebra to analyze this behavior for interesting features.