posted by H .
A ball is dropped from the top of a building. The balls fall 3 times farther during its last second of freefall than it did during its third second of freefall. The ball fell 4.9 meters during its first second, how tall is the building and how much time is the ball in the air?
Answer d = 313.6 m
distance fallen: s(t) = 4.9t^2
during 3rd second, it fell 4.9(3^2-2^2) = 24.5m
so, during its final second, it fell 73.5m
73.5 = 4.9 (t^2 - (t-1)^2)
t = 8
so, the building height is 4.9*8^2 = 313.6