A particle dropped from vertically downward from the top of the building, find the displacement of the nth second, 1st, 2nd, 3rd and 4th seconds.

Answer 1

The distance fallen after t seconds is

s(t) = 4.9t^2 meters

So just plug in values for t. If you are interested in the distance fallen during the nth second, then just subtract the previous position. That distance d(t) is then just

d(t) = s(t) - s(t-1)