Respuesta :
Answer:
The rock hits the ground at t = 4.68 seconds.
Step-by-step explanation:
The height of the rock after t seconds is given by the following equation:
[tex]h(t) = -16t^{2} + v(0)t + h(0)[/tex]
In which v(0) is the initial speed and h(0) is the initial height.
An rock is thrown downward from a platform that is 163 feet above ground at 40 feet per second.
This means that [tex]v(0) = 40, h(0) = 163[/tex]
So
[tex]h(t) = -16t^{2} + 40t + 163[/tex]
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]
When the rock hits the ground?
When h(t) = 0. So
[tex]h(t) = -16t^{2} + 40t + 163[/tex]
[tex]-16t^{2} + 40t + 163 = 0[/tex]
So
[tex]a = -16, b = 40, c = 163[/tex]
Then
[tex]\bigtriangleup = (40)^{2} - 4*(-16)*163 = 12032[/tex]
[tex]t_{1} = \frac{-40 + \sqrt{12032}}{2*(-16)} = -2.18[/tex]
[tex]t_{2} = \frac{-40 - \sqrt{12032}}{2*(-16)} = 4.68[/tex]
Time is a positive measure, then:
The rock hits the ground at t = 4.68 seconds.
