# 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

*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 [latex]4[/latex]-foot pedestal. The height (*h,*in feet) of the rocket

*t*seconds after taking off is given by the formula [latex]h=−2t^{2}+7t+4[/latex]. 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 [latex]0[/latex]. We want to know how long, t, the rocket is in the air.

**Translate:**So, we will substitute [latex]0[/latex] for

*h*in the formula and solve for t.

[latex]\begin{array}{l}h=−2t^{2}+7t+4\\0=−2t^{2}+7t+4\end{array}[/latex]

**Write and Solve:**Rewrite the middle term using the [latex]a\cdot{c}[/latex] method.

[latex]0=-2t^{2}+8t-t+4[/latex]

Factor the trinomial by grouping.[latex]\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}[/latex]

Use the Zero Product Property. There is no need to set the constant factor [latex]-1[/latex] to zero, because [latex]-1[/latex] will never equal zero.[latex]2t+1=0\,\,\,\,\,\,\text{or}\,\,\,\,\,\,t-4=0[/latex]

Solve each equation.[latex]t=-\frac{1}{2}\,\,\,\,\,\,\text{or}\,\,\,\,\,\,t=4[/latex]

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

#### Answer

The rocket will hit the ground [latex]4[/latex] seconds after being launched.### Example

Use the formula for the height of the rocket in the previous example to find the time when the rocket is [latex]4[/latex] 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 [latex]4[/latex] 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 = [latex]4[/latex] 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:**

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

Now we can factor out a t from each term:

[latex]0=t\left(-2t+7\right)[/latex]

Solve each equation for t using the zero product principle:

[latex]\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}[/latex]

**Interpret: **It doesn't make sense for us to choose t=[latex]0[/latex] 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=[latex]3.5[/latex]

#### Answer

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

## Licenses & Attributions

### CC licensed content, Original

- Parabolic motion description and example.
**Provided by:**Lumen Learning**License:**CC BY: Attribution. - Factoring Application - Find the Time When a Projectile Hits and Ground.
**Authored by:**James Sousa (Mathispower4u.com) for Lumen Learning.**License:**CC BY: Attribution.

### CC licensed content, Shared previously

- Parabolic water trajectory.
**Authored by:**By GuidoB.**Located at:**https://commons.wikimedia.org/w/index.php?curid=8015696.**License:**CC BY-SA: Attribution-ShareAlike.