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=−2t2+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.
h=−2t2+7t+40=−2t2+7t+4
Write and Solve: Rewrite the middle term using the
a⋅c method.
0=−2t2+8t−t+4
Factor the trinomial by grouping.
0=−2t(t−4)−(t−4)0=(−2t−1)(t−4)0=−1(2t+1)(t−4)
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=0ort−4=0
Solve each equation.
t=−21ort=4
Interpret the answer. Since
t represents time, it cannot be a negative number; only
t=4 makes sense in this context.
t=4
Answer
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.
h=−2t2+7t+4
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:
h=−2t2+7t+44=−2t2+7t+4−4−40=−2t2+7t
Now we can factor out a t from each term:
0=t(−2t+7)
Solve each equation for t using the zero product principle:
t=0 OR −2t+7=0−7−7−2−2t=−2−7t=27=3.5
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
Answer
t=3.5 seconds
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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