A roofing tile falls from rest off the roof of a building. An observer from across the street notices that it takes 0.32 s for the tile to pass between two windowsills that are 3.49 m apart. How far is the sill of the upper window from the roof of the building?
The tile falls with a constant acceleration of g, at the upper sill it will have achieved some velocity V What we know that if falls 3.49 m in a time of 0.32 sec, starting with the velocity V and accelerating at g d = Vt + 0.5gt^2 3.49 = 0.32V + 4.9(0.32)^2 V = 9.34 m/s Now just find how far something would have to fall to achieve a speed of 9.34 m/s using Vf^2 = Vi^2 + 2ad Vf = 9.34 m/s, Vi = 0 m/s , a = g = 9.8 m/s^2 9.34^2 = 2(9.8)d d = 4.45 m from the roof
Called Vx the speed the tile has reached in free falling mode at the upper window's sill span = 3.49 = (Vx+(Vx+g*t)/2*t 6.98 = 2Vx*0.32+9.8*0.32^2 5.976 = 0.64Vx Vx = 5.976/0.64 = 9.34 m/sec (tile speed at the sill of the upper window) falling time tf = Vx/g = 9.34/9.8 = 0.953 sec (from roof to the sill of the upper window) h = 1/2gt^2 = 4.9*0.953^2 = 4.45 m h = Vx^2/2g = 9.34^2/19.6 = 4.45 m
