Posted by danielle on Sunday, March 25, 2012 at 8:51pm.
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?

Math  Damon, Sunday, March 25, 2012 at 9:32pm
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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