+108 I am posting this question again, here is a link to the original question.
http://web2.0calc.com/questions/mechanics_2
According to the text book, the answer is 20.4m.
A steel ball is dropped from a building's roof and passes a window, taking 0.125 s to fall from the top to the bottom of the window, a distance of 1.20m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.125 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is 2.0 s. How tall is the building?
+36916 I do not see my post to this question, so I will repost:
From FIRST post we found
Speed at top of window = 8.9875 m/s Distance from roof top to top of window 4.12 m
NOW
Time to travel past window = .125 then to ground 1 sec = total 1.25 sec
x = distance to top of window from ground
= xo+vot+1/2 at^2 where xo=0 vo= 8.9875 m/s (NOT zero as I previously used) t= 1.125
= 8.9875 (1.125) + 1/2(9.8)(1.125^2) = 16.31m
16.31 m <-------- add THIS to the distance from the top of the building to top of window
16.31 + 4.12 = 20.43 m = building height (sorry for the previous error)
