Question 1092505
the height of the object at time t is modeled by the following formula
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s(t) = –gt^2 + v0t + h0, where g is the acceleration due to gravity, v0 is the objects initial velocity, h0 is the initial height of the object
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since we are working in feet, g=16, also v0=110 and h0=75
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s(t) = -16t^2 +110t +75
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this is a parabola that curves downward
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the graph of this equation is 
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{{{ graph( 300, 200, -2, 9, -100, 300, -16x^2+110x+75) }}} 
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a)  120 = -16t^2 +110t +75
-16t^2 +110t -45 = 0
t^2 -6.875t +2.8125 = 0
use quadratic formula to solve for t
t = (-(-6.875) +square root((-6.875)^2 -4(1)(2.8125))) / 2(1) = 6.4382 seconds
t = (-(-6.875) -square root((-6.875)^2 -4(1)(2.8125))) / 2(1) = 0.4368 seconds
Note that the object attains the height of 120 feet at two different times(on the way up and again on the way down)
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b)  t = -b/2a(this is the t value associated with the vertex)
s(t) = -16t^2 +110t +75
t = -110 / 2(-16) = 3.4375
s(3.4375) = -16(3.4375)^2 +110(3.4375) +75 = 264.0625 feet at its maximum height
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c)  s is 0 when the object hits the ground
0 = -16t^2 +110t +75
t^2 -6.875t -4.6875 = 0
t = (-(-6.875) +square root((-6.875)^2 -4(1)(-4.6875))) / 2(1) = 7.5
t = (-(-6.875) +square root((-6.875)^2 -4(1)(-4.6875))) / 2(1) = -0.625
we reject the negative value for t, therefore the object hits the ground after 7.5 seconds
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