Respuesta :
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:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]\bigtriangleup = b^{2} - 4ac[/tex]
The height of the bullet after t seconds is given by:
[tex]h(t) = -16t^2 + 720t[/tex]
When does the bullet reach a height of 4,000 feet?
This is t for which [tex]h(t) = 4000[/tex]. So
[tex]4000 = -16t^2 + 720t[/tex]
[tex]16t^2 - 720t + 4000 = 0[/tex]
Dividing by 16
[tex]t^2 - 45t + 250 = 0[/tex]
So [tex]a = 1, b = -45, c = 250[/tex]
[tex]\bigtriangleup = b^{2} - 4ac = (-45)^2 - 4(1)(250) = 1025[/tex]
[tex]t_{1} = \frac{-(-45) + \sqrt{1025}}{2} = 38.5[/tex]
[tex]t_{2} = \frac{-(-45) - \sqrt{1025}}{2} = 6.49[/tex]
The bullet reaches a height of 4000 feet after 6.49 seconds, and then, coming back down, after 38.5 seconds.
