Answer:
D) 484 feet
Step-by-step explanation:
Given: [tex]H(t) = -16t^2 + vt + s[/tex], where t is the time, v is the initial velocity/speed and s is initial height.
and
v = 144 feet and s = 160 feet
Now plug in the given values in the given [tex]H(t) = -16t^2 + vt + s[/tex]
[tex]H(t) = -16t^2 + 144t + 160[/tex]
This is in the form of quadratic function [tex]f(x) = ax^2 + bx + c, where a \neq 0[/tex]
The maximum height attain in vertex. Here we have formula to find the vertex of the x coordinate.
x = [tex]\frac{-b}{2a}[/tex]
Here b = 144 and a = -16
So, t = [tex]\frac{-144}{2*-16} = \frac{-144}{-32} = 4.5[/tex]
The maximum height attain at t = 4.5 seconds. To find the maximum height plug in t = 4.5 in [tex]H(t) = -16t^2 + 144t + 160[/tex]
H(4.5) = [tex]-16(4.5)^2 + 144 (4.5) + 160\\[/tex]
= -16(20.25) + 648 + 160
= -324 + 648 + 160
= -324 + 808
= 484 feet
So the maximum height of the particle is 484 feet.
