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Question 993171: Please help I do not understand any of these questions. These are all quadratic formulas. Please show all work it helps me learn how to do the questions step by step.
1. Imagine yourself standing on the roof of the 1450-foot Sears Tower in Chicago. When you throw a baseball upward from the roof of the tower, the ball’s height above the ground, H (in feet), can be described as a function of the time, t (in seconds), since the ball was dropped. This height function is defined by
H(t) = -16t2 + 80t +1450
a. Identify the vertical intercept and what does it mean in the context of the problem
b. After the baseball is thrown, how long will the ball hit the ground?
c. Your answer to question #2 identifies one horizontal intercept of the equation. Identify another horizontal intercept and discuss whether it makes sense in the context of the problem.
d. Find the vertex of the equation and interpret it meaning in the context of the problem.
e. How far from the ground is the ball after 8 seconds?
f. At what time(s) is the ball 1500 feet above the ground?
g. Identify the practical domain in the context of the problem.
h. Identify the practical range in the context of the problem.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! The equation H(t) = -16t^2 + 80t +1450 describes a parabola that opens downward
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a) at time t = 0, H(0) = 1450, 1450 is the vertical intercept and says that the ball is 1450 feet above the ground at time t = 0
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b) we must solve the quadratic for t when H(t) = 0, that is
-16t^2 + 80t +1450 = 0
divide both sides of = by -2
8t^2 - 40t - 725 = 0
use quadratic formula to solve for t
t = (40 + sqrt((-40^2) - 4*8*(-725)))/(2*8) = 12.342509843 approx 12.343
t = (40 - sqrt((-40^2) - 4*8*(-725)))/(2*8) = −7.342509843 approx -7.343
it takes 12.343 seconds before the ball hits the ground
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c) the other horizontal intercept is -7.343, we can not have negative time
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d) the t for the vertex is given by the following formula
t = -b / 2a, then
t = -80 / 2(-16) = 2.5
now substitute t = 2.5 in the H(t) equation
H(t) = -16t^2 + 80t +1450
H(2.5) = -16(2.5)^2 + 80(2.5) + 1450
H(2.5) = 1550
the vertex is (2.5, 1550), this tells us that at time 2.5 seconds the ball reaches its maximum height of 1550 feet
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e) H(8) = -16(8)^2 + 80(8) + 1450
H(8) = 1066
after 8 seconds the ball is 1066 feet from the ground
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f) at time 2.5 seconds the ball is 1550 feet above the ground
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g) practical domain is t values from 0 to 12.343
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h) practical range is height values from 0 to 1550
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