# MAT 150

A projectile was launched from ground level with an initial velocity of the v0 neglecting air resistance is high in feet two seconds after launch is given by S= -16T^2 + v0t find the times that the projectile will reach a high of 128 feet and return to the ground when the zero equals 96 feet per second

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1. what does "the zero equals 96 feet per second" mean?
I think you meant to say v0 = 96 ft/s
s = -16t^2 + 96t
so you need to solve
-16t^2 + 96t = 128
t^2 - 6t + 8 = 0
(t-2)(t-4) = 0
t=2 (on the way up)
t=4 (on the way back down)
and
-16t^2 + 96t = 0
-16t(t-6) = 0
t=0 (at the start)
t=6 (upon impact)

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