Calculus
posted by Anonymous on .
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