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: Find when the bullet hits the ground

The bullet hits the ground when $h = 0$. So we set the height - equation $h=-16t^{2}+720t = 0$.
Factor out $t$: $t(-16t + 720)=0$.
This gives two solutions for $t$: $t = 0$ (corresponds to the time of firing) and $-16t+720 = 0$.
Solve $-16t+720 = 0$ for $t$.
Add $16t$ to both sides: $720=16t$.
Then $t=\frac{720}{16}=45$ seconds.

Step2: Find when the bullet reaches a height of 4000 feet

Set $h=-16t^{2}+720t = 4000$.
Rearrange to get a quadratic equation in standard form: $16t^{2}-720t + 4000=0$.
Divide through by 16: $t^{2}-45t + 250=0$.
Use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $at^{2}+bt + c = 0$. Here, $a = 1$, $b=-45$, and $c = 250$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-45)^{2}-4\times1\times250=2025 - 1000=1025$.
Then $t=\frac{45\pm\sqrt{1025}}{2}=\frac{45\pm5\sqrt{41}}{2}$.
$t_1=\frac{45 + 5\sqrt{41}}{2}\approx\frac{45+5\times6.4}{2}=\frac{45 + 32}{2}=\frac{77}{2}=38.5$ and $t_2=\frac{45 - 5\sqrt{41}}{2}\approx\frac{45-32}{2}=\frac{13}{2}=6.5$ seconds.

Answer:

1. 45 seconds
2. 6.5 seconds and 38.5 seconds