# Calculus

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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?

• Calculus -

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

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