A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s(t)=96t16t^2
a.) at what time will the ball strike the ground
b.) for what time t is the ball more than 128 feet above the ground?
c.) when will the ball reach its highest peak? how high is it above the ground?
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1 answer

a)
when is s = 0 ?
0 = t(96  16 t)
t = 0 of course, it starts at 0
t = 96/16 = 6 seconds
b)
when is s = 128 (two times, on the way up and on the way down)
128 = 96 t  16 t^2
8 = 6 t  t^2
t^2  6 t + 8 = 0
(t4)(t2) = 0
so between 2 seconds and 4 seconds
c)where is the vertex of this parabola?
Well it is halfway in time between 2 seconds and four seconds, which is 3 seconds
it is also halfway in time between 0 seconds and 6 seconds, which of course is also at 3 seconds
so
s(3) = 96(3)  16(9)
= 144 feet 👍
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answered by Damon
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