When does the bullet fall back to the ground?

Question

Abullet is fired straight upward with an initial speed of 720 ft/s. It's path is modeled by the equation h=-16t²+720t, where h is the height of the bullet t seconds after it wad fired.

Answer 1

This is quite a simple one, all you have to do is find the intercepts of the bullet's path equation through the time axis.

Let's first ask ourselves at what height the bullet is at when it reaches the ground. If we define the height of the bullet to be the distance from the bullet to the ground, then we can say that the height of the bullet when it hits the ground is 0 ft. Now that we know that, we need to solve for the time it hits the ground. Let's say that time when the bullet is fired is 0 seconds and the bullet hits the ground at some time t_f.

h(t_f) = 0 = -16t_f² + 720t_f

= (-16)(t_f² - 45t_f)

Dividing both sides by -16, we get

0 = (t_f² - 45t_f)

= (t_f) * (t_f - 45)

Solving for t_f, we get two possible solutions, 0 seconds and 45 seconds. The first solution refers to the time at which the bullet was first fired, which we defined 0 seconds on our clock. Therefore, the alternative answer must be when the bullet reaches the ground again during its trip. So the answers is that the bullet hits the ground 45 seconds after it was fired.