SOLUTION: An object is projected vertically from the top of a building with an initial velocity of 128ft/sec. It's distance is s(t) in feet above the ground after t seconds is given by the e

Question 946175: An object is projected vertically from the top of a building with an initial velocity of 128ft/sec. It's distance is s(t) in feet above the ground after t seconds is given by the equation.
S(t)= -16t^2 + 128t +80
a. Find its maximum distance above the ground.
b. Find the height of the building.

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
An object is projected vertically from the top of a building with an initial velocity of 128ft/sec. It's [sic] distance is s(t) in feet above the ground after t seconds is given by the equation.
S(t)= -16t^2 + 128t +80
a. Find its maximum distance above the ground.
Max height is the vertex of the parabola at t - -b/2a
t = -128/-32 = 4 seconds
s(4) = -16*16 + 128*4 + 80 = 336 feet
=======================
b. Find the height of the building.
It's s(0) = 80 feet
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