Question 212103
A soccer ball is kicked straight up with an initial velocity of 32 ft. per second. Its height above the earth is given by s (t) = -16t2 + 32t, where,‘s’ is the height in ft. at any time, ‘t’ seconds.
What is the maximum height reached by the ball?
How long after it was thrown up the ball reaches its maximum height? 
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The "vertex" of the parabola will give you both.
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First, find the "axis of symmetry":
t = -b/2a
t = -32/2(-16)
t = -32/-32
t = 1 second (answer to second question)
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The "height" is found by plugging the value above back into:
s(t) = -16t^2 + 32t
s(1) = -16(1)^2 + 32(1)
s(1) = -16 + 32
s(1) = 16 feet (answer to first question)