Answer:
a) s = -16 + 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 +
t +
where: s is the height of the football above the ground, t is the time, is the initial velocity and
is the initial height.
The height of the building = = 192 feet and the initial speed of the ball =
= 96 feet per second, then;
s = -16 + 192t + 96
a) The polynomial is given as:
s = -16 + 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.
Answer:
31.4 in³
Step-by-step explanation:
The box is just big enough to hold the 3 balls, so it must have a length 6 times the radius of each ball, a width 2 times the radius, and a height 2 times the radius.
The volume of the box is:
V = (6r)(2r)(2r)
V = 24r³
The volume of the 3 balls is:
V = 3 (4/3 π r³)
V = 4πr³
So the volume of the air is:
V = 24r³ − 4πr³
V = (24 − 4π) r³
Since r = 1.4 inches:
V = (24 − 4π) (1.4 in)³
V ≈ 31.4 in³
