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
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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.
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
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b. Find the height of the building.
It's s(0) = 80 feet
