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# Projectiles

### Learning Outcomes

• Define projectile motion
• Solve a quadratic equation that represents projectile motion
• Interpret the solution to a quadratic equation that represents projectile motion
Projectile motion happens when you throw a ball into the air and it comes back down because of gravity.  A projectile will follow a curved path that behaves in a predictable way.  This predictable motion has been studied for centuries, and in simple cases it's height from the ground at a given time, t, can be modeled with a quadratic polynomial of the form $\text{height}=at^2+bt+c$ such as we have been studying in this module. Projectile motion is also called a parabolic trajectory because of the shape of the path of a projectile's motion, as in the image of water in the fountain below.
Parabolic WaterTrajectory
Parabolic motion and it's related equations allow us to launch satellites for telecommunications, and rockets for space exploration. Recently, police departments have even begun using projectiles with GPS to track fleeing suspects in vehicles, rather than pursuing them by high-speed chase [footnote]"Cops' Latest Tool in High-speed Chases: GPS Projectiles." CBSNews. CBS Interactive, n.d. Web. 14 June 2016.[/footnote]. In this section we will solve simple quadratic polynomials that represent the parabolic motion of a projectile. The real mathematical model for the path of a rocket or a police GPS projectile may have different coefficients or more variables, but the concept remains the same. We will also learn to interpret the meaning of the variables in a polynomial that models projectile motion.

### Example

A small toy rocket is launched from a $4$-foot pedestal. The height (h, in feet) of the rocket t seconds after taking off is given by the formula $h=−2t^{2}+7t+4$. How long will it take the rocket to hit the ground?

Answer: Read and understand: The rocket will be on the ground when the height is $0$. We want to know how long, t,  the rocket is in the air. Translate: So, we will substitute $0$ for h in the formula and solve for t.

$\begin{array}{l}h=−2t^{2}+7t+4\\0=−2t^{2}+7t+4\end{array}$

Write and Solve: Rewrite the middle term using the $a\cdot{c}$ method.

$0=-2t^{2}+8t-t+4$

Factor the trinomial by grouping.

$\begin{array}{l}0=-2t\left(t-4\right)-\left(t-4\right)\\0=\left(-2t-1\right)\left(t-4\right)\\0=-1\left(2t+1\right)\left(t-4\right)\end{array}$

Use the Zero Product Property. There is no need to set the constant factor $-1$ to zero, because $-1$ will never equal zero.

$2t+1=0\,\,\,\,\,\,\text{or}\,\,\,\,\,\,t-4=0$

Solve each equation.

$t=-\frac{1}{2}\,\,\,\,\,\,\text{or}\,\,\,\,\,\,t=4$

Interpret the answer. Since t represents time, it cannot be a negative number; only $t=4$ makes sense in this context. [latex-display]t=4[/latex-display]

The rocket will hit the ground $4$ seconds after being launched.

In the next example we will solve for the time that the rocket is at a given height other than zero.

### Example

Use the formula for the height of the rocket in the previous example to find the time when the rocket is $4$ feet from hitting the ground on it's way back down.  Refer to the image. [latex-display]h=−2t^{2}+7t+4[/latex-display]

Answer: Read and understand: We are given that the height of the rocket is $4$ feet from the ground on it's way back down. We want to know how long it has taken the rocket to get to that point in it's path, we are going to solve for t. Translate: Substitute h = $4$ into the formula for height, and try to get zero on one side since we know we can use the zero product principle to solve polynomials. Write and Solve:

$\begin{array}{l}h=−2t^{2}+7t+4\\4=-2t^2+7t+4\\\underline{-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-4}\\0=-2t^2+7t\end{array}$

Now we can factor out a t from each term:

$0=t\left(-2t+7\right)$

Solve each equation for t using the zero product principle:

$\begin{array}{l}t=0\text{ OR }-2t+7=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-7}\,\,\,\,\,\,\,\underline{-7}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{-2t}{-2}=\frac{-7}{-2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=\frac{7}{2}=3.5\end{array}$

Interpret: It doesn't make sense for us to choose t=$0$ because we are interested in the amount of time that has passed when the projectile is 4 feet from hitting the ground on it's way back down. We will choose t=$3.5$

[latex-display]t=3.5\text{ seconds }[/latex-display]

The video that follows presents another example of solving a quadratic equation that represents parabolic motion. https://youtu.be/hsWSzu3KcPU In this section we introduced the concept of projectile motion, and showed that it can be modeled with a quadratic polynomial.  While the models used in these examples are simple, the concepts and interpretations are the same.  The methods used to solve quadratic polynomials that don't factor easily are many and well known, it is likely you will come across more in your studies.

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