Question

use the following information to answer questions 17 - 18: a bullet is fired straight upward with an initial speed of 720 ft/s. its path is modeled by the equation h = - 16t²+ 720t, where h is the height of the bullet t seconds after it was fired. 17. when does the bullet fall back to the ground? 18. when does the bullet reach a height of 4,000 feet?

Explanation:

Step1: Set up the height - equation for question 17

The height of the bullet is given by the equation $h=-16t^{2}+720t$. When the bullet falls back to the ground, $h = 0$. So we set up the equation $-16t^{2}+720t=0$.

Step2: Factor out the common factor

Factor out $-16t$ from the left - hand side of the equation: $-16t(t - 45)=0$.

Step3: Solve for $t$

Using the zero - product property, if $ab = 0$, then either $a = 0$ or $b = 0$. So $-16t=0$ gives $t = 0$ (this is the time of firing), and $t-45=0$ gives $t = 45$ seconds.

Step4: Set up the height - equation for question 18

We want to find when $h = 4000$. So we set up the equation $-16t^{2}+720t=4000$.

Step5: Rearrange the equation to standard quadratic form

Rearrange it to $16t^{2}-720t + 4000=0$. Divide through by 16 to simplify: $t^{2}-45t + 250=0$.

Step6: Solve the quadratic equation

We can solve the quadratic equation $t^{2}-45t + 250=0$ using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a = 1$, $b=-45$, and $c = 250$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-45)^{2}-4\times1\times250=2025 - 1000=1025$. Then $t=\frac{45\pm\sqrt{1025}}{2}=\frac{45\pm5\sqrt{41}}{2}$. So $t=\frac{45 + 5\sqrt{41}}{2}\approx\frac{45+5\times6.4}{2}=\frac{45 + 32}{2}=\frac{77}{2}=38.5$ or $t=\frac{45 - 5\sqrt{41}}{2}\approx\frac{45-32}{2}=\frac{13}{2}=6.5$ seconds.

Answer:

1. 45 seconds
2. $t=\frac{45 + 5\sqrt{41}}{2}\approx38.5$ seconds or $t=\frac{45 - 5\sqrt{41}}{2}\approx6.5$ seconds