Question

A bullet is fired straight upward with an initial speed of 720 ft/s. It’s path is modeled by the equation h=-16t^2 + 720t, where h is the height of the bullet t seconds after it was fired. When does the bullet reach a height of 4,000 feet?

Answer 1

Answer:

The bullet reaches a height of 4000 feet after 6.49 seconds, and then, coming back down, after 38.5 seconds.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

The height of the bullet after t seconds is given by:

$h(t) = -16t^2 + 720t$

When does the bullet reach a height of 4,000 feet?

This is t for which $h(t) = 4000$. So

$4000 = -16t^2 + 720t$

$16t^2 - 720t + 4000 = 0$

Dividing by 16

$t^2 - 45t + 250 = 0$

So $a = 1, b = -45, c = 250$

$\bigtriangleup = b^{2} - 4ac = (-45)^2 - 4(1)(250) = 1025$

$t_{1} = \frac{-(-45) + \sqrt{1025}}{2} = 38.5$

$t_{2} = \frac{-(-45) - \sqrt{1025}}{2} = 6.49$

The bullet reaches a height of 4000 feet after 6.49 seconds, and then, coming back down, after 38.5 seconds.