Question 993171
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