Asked by Phy_looser
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.
Answers
Answered by
Steve
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)
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)
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