v0 t - 16t^2 = 0 when t=8, so
8 v0 - 1024 = 0
v0 = 128
Now just find the vertex of the parabola, at t = -b/2a
which in this case is at t = v0/32 = 4
that sounds correct, since it is at half the time to hit. 4 sec up, 4 sec down. So, just find s(4)
An object is projected vertically upward with an initial velocity of v0 ft/sec, and its distance s(t) in feet above the ground after t seconds is given by the formula s(t) = −16t^2 + v0t.
(a) If the object hits the ground after 8 seconds, find its initial velocity v0.
v0 = ft/sec
(b) Find its maximum distance above the ground.
s(t) = ft
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