Respuesta :
Answer:
(a) The bridge's height above the water is 114 feet.
(b) After 4.332 seconds the rock hit the water.
(c) After 1.34375 seconds the rock reach its maximum height above the water.
(d) The rock's maximum height above the water is 142.89 feet.
Step-by-step explanation:
The given function is
[tex]f(t)=-16t^2+43t+114[/tex]
(a)
At t=0
[tex]f(0)=-16(0)^2+43(0)+114=114[/tex]
Therefore, the bridge's height above the water is 114 feet.
(b)
If a quadratic equation is defined as [tex]ax^2+bx+c=0[/tex], then the quadratic formula is
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
In the given function a=-16, b=43 and c=114.
[tex]t=\dfrac{-43\pm \sqrt{(43)^2-4(-16)(114)}}{2(-16)}[/tex]
[tex]t=\dfrac{-43+\sqrt{(43)^2-4(-16)(114)}}{2(-16)},\dfrac{-43-\sqrt{(43)^2-4(-16)(114)}}{2(-16)}[/tex]
[tex]t=-1.645,4.332[/tex]
Time can not be negative. It means, at t=4.332 the rock hit the water.
Therefore, after 4.332 seconds the rock hit the water.
(c)
The vertex of quadratic function [tex]f(t)=at^2+bt+c[/tex] is
[tex]vertex (-\frac{b}{2a},f(-\frac{b}{2a}))}[/tex]
For the given function a=-16, b=43 and c=114.
The leading coefficient is negative. It means the vertex is point of maxima.
[tex]-\frac{b}{2a}=-\frac{43}{2(-16)}=1.34375[/tex]
Therefore, after 1.34375 seconds the rock reach its maximum height above the water.
(d)
Substitute t=1.34375 in the given function.
[tex]f(1.34375)=-16(1.34375)^2+43(1.34375)+114=142.89[/tex]
Therefore, the rock's maximum height above the water is 142.89 feet.
From the quadratic equation, we get that:
a) The bridge is 114 feet above water.
b) The rock hits the water 4.33 seconds after being thrown.
c) The rock reaches it's maximum height 1.34 seconds after being thrown.
d) The rock's maximum height is of 142.9 feet above the water.
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The height of the rock after t seconds is given by:
[tex]f(t) = -16t^2 + 43t + 114[/tex]
Which is a quadratic equation with [tex]a = -16, b = 43, c = 114[/tex].
Item a:
- The rock was thrown from the bridge, thus it's height is f(0).
[tex]f(0) = -16(0)^2 + 43(0) + 114 = 0[/tex]
The bridge is 114 feet above water.
Item b:
- It hits the water when f(t) = 0, thus, we have to solve the quadratic equation.
[tex]-16t^2 + 43t + 114 = 0[/tex]
[tex]\Delta = 43^2 - 4(-16)(114) = 9145[/tex]
[tex]t_{1} = \frac{-43 + \sqrt{9145}}{2(-16)} = -1.64[/tex]
[tex]t_{2} = \frac{-43 - \sqrt{9145}}{2(-16)} = 4.33[/tex]
We want the positive value, so:
The rock hits the water 4.33 seconds after being thrown.
Item c:
- This is the t-value of the vertex, which is:
[tex]t_v = -\frac{b}{2a}[/tex]
So
[tex]t_v = -\frac{43}{2(-16)} = \frac{43}{32} = 1.34[/tex]
The rock reaches it's maximum height 1.34 seconds after being thrown.
Item d:
- This is the f-value of the vertex, which is:
[tex]f_v = -\frac{\Delta}{4a}[/tex]
So
[tex]t_v = -\frac{9145}{4(-16)} = \frac{9145}{64} = 142.9[/tex]
The rock's maximum height is of 142.9 feet above the water.
A similar problem is given at https://brainly.com/question/24737967
