A roofing tile falls from rest off the roof of a building. An observer from across the street notices that it takes 0.37 s for the tile to pass between two windowsills that are 2.5 m apart. How far is the sill of the upper window from the roof of the building?
V1 = 0 g = 9.8 t = 0.40 9 a million) locate D, distance from the roof to the bottom window, utilizing D = v1*t + 0.5 * a * (t^2), a = g ! 2) subtract 2.6 out of your D value, to get distance between roof and top window, and thats your answer !
The average speed as it passed from one windowsill to the other is given by Vave = 2.5m / 0.91 m The instantaneous speed when passing the midpoint of that 2.5m would be equal to Vave. The length of fall required to reach a speed of Vave (which it had at the midpoint) is given by the kinematic formula Vf^2 = Vo^2 + 2*a*d where Vf is instantaneous speed when passing the midpoint, in other words Vf = Vave; Vo = 0 (it was dropped, not thrown down; a = g, and d is how far that midpoint is from the roof. So the top windowsill is 1/2 of the 2.5 m closer to the top.
