An object is shot straight down from the top of a 220 ft building with an initial velocity of -22 ft/sec. What is the position function? What is the velocity of the object after it has fallen 108 ft?

Answer 1

1)

Position, s(t) = gt²/2+v₀t+s₀

For this case:

g = -32 ft/sec²

v₀ = -22 ft/sec

s₀ = 220 ft

s(t) = -16t² - 22t + 220

2)

First, need to solve for t when s(t) = 108

108 = -16t²-22t+220

-16t² -22t + 112 = 0

By quadratic equation, t = 2, -27/8 Choose t = 2

v(t) = gt+v₀ = -32t-22

v(2) = -64-22 = -86 ft/sec