SOLUTION: An object is projected vertically upward from the top of a building with an initial velocity of 112 ft/sec. Its distance in feet above the ground after t seconds is given by the eq

Question 364549: An object is projected vertically upward from the top of a building with an initial velocity of 112 ft/sec. Its distance in feet above the ground after t seconds is given by the equation

s (t) = -16t^2 + 112t + 87
Find its maximum distance above the ground
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
s(t) = -16t^2 + 112t + 87
by inspection (examination of coefficient associated with the t^2 term), we see that it is a parabola that opens downward (sad face).
.
The "vertex" will be give you the max height.
Time at max height:
t = -b/(2a) = -112/(2*(-16)) = -112/(-32) = 3.5 secs
.
To find height, plug it into the equation:
s(t) = -16t^2 + 112t + 87
s(3.5) = -16(3.5)^2 + 112(3.5) + 87
s(3.5) = 283 feet

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