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)
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.