posted by Anonymous .
A fireworks shell is shot upward with an initial velocity of 28m/s from a height of 2.5 m.
s(t) = -4.9t^2 + 28t + 2.5
After how many seconds will the shell have the same speed, but be falling downward?
v(t) = -9.8t + 28
we want this to be -28 (negative for downwards)
-9.8t + 28 = -28
-9.8t = -56
t = -56/-9.8 = 5.7 seconds