Question

pyrotechnicians can use polynomials to plan complex fireworks displays. a firework is launched from a platform 6 feet above the ground at a speed of 200 feet per second. the firework has a 5 - second fuse. the height of the firework in feet is given by the polynomial (-16t^2 + 200t + 6), where (t) is the time in seconds. how high will the firework be when it explodes? after you solved this problem, create another problem using a different scenario that also utilized a polynomial equation.

Explanation:

Step1: Identify the time when it explodes

The firework has a 5 - second fuse, so \( t = 5 \) seconds when it explodes.

Step2: Substitute \( t = 5 \) into the height polynomial

The height polynomial is \( h(t)=- 16t^{2}+200t + 6 \). Substitute \( t = 5 \) into the polynomial:
\[

$$\begin{align*} h(5)&=-16\times(5)^{2}+200\times5 + 6\\ &=-16\times25+1000 + 6\\ &=-400 + 1000+6\\ &=600 + 6\\ &=606 \end{align*}$$

\]

Answer:

The firework will be 606 feet high when it explodes.

For creating another problem:
Problem: A ball is thrown from a height of 4 feet above the ground with an initial velocity of 150 feet per second. The height \( h \) of the ball in feet after \( t \) seconds is given by the polynomial \( h(t)=-16t^{2}+150t + 4 \). How high will the ball be after 3 seconds?

Solution:

Step1: Identify the time

We need to find the height at \( t = 3 \) seconds.

Step2: Substitute \( t = 3 \) into the height polynomial

\[

$$\begin{align*} h(3)&=-16\times(3)^{2}+150\times3 + 4\\ &=-16\times9+450 + 4\\ &=-144+450 + 4\\ &=306 + 4\\ &=310 \end{align*}$$

\]

So the ball will be 310 feet high after 3 seconds.