Answer:
a) s = -16[tex]t^{2}[/tex] + 192t + 96
b) Height of the football after 2 seconds is 448 feet.
ii. The football would not hit the ground after 2 seconds.
Step-by-step explanation:
Given:
s = -16[tex]t^{2}[/tex] + [tex]V_{o}[/tex]t + [tex]s_{o}[/tex]
where: s is the height of the football above the ground, t is the time, [tex]V_{o}[/tex] is the initial velocity and [tex]s_{o}[/tex] is the initial height.
The height of the building = [tex]V_{o}[/tex] = 192 feet and the initial speed of the ball = [tex]s_{o}[/tex] = 96 feet per second, then;
s = -16[tex]t^{2}[/tex] + 192t + 96
a) The polynomial is given as:
s = -16[tex]t^{2}[/tex] + 192t + 96
b) The height of the football when t = 2 seconds is;
s = -16(2) + 192(2) + 96
 = -32 + 384 + 96
 = -32 + 480
s = 448
The height of the football when t = 2 seconds is 448 feet.
ii. The football would not hit the ground after 2 seconds.
