A roof tile falls off the roof of a house. An observer inside the house notices that the tile took 0.2 s to pass a 1.6 m window of the house. If the tile feel from rest, how far above the top of the window did it fall from? Give your answer in meters, recall that this is height above the window and has a positive sign, do not enter units.
1.6 m/0.2 secs = average velocity = 8 m/s 8 - ((0.2/2) x 9.8) = 7.02 m/s v = at 7.02/a = 7.02/9.8 = t = 0.7163 secs t^2 x (1/2a) = t^2 x 4.9 = 2.5143 m
first s= ut+1/2gt^2 1.6=0.2u+0.5*10*0.04 1.6=0.2(u+1) u+1=8 u=7 height = (7^2)/(2*10) = 2.45 using third eqn of kinematics
Average velocity of tile in falling past window = 1.6/0.2 = 8m/s This is the instantaneous velocity of the tile half way down the window in time. This means that the tile had a velocity of 8m/s 0.1 sec after it reached the top of the window. How long does it take a falling body to attain velocity of 8m/s, v = u +at 8 = 0 + 9.8t t= 8/9.8 t = 0.816sec 0.816 - 0.1 = 0.716 seconds Therefore it reached the top of the window in 0.716 seconds. How far must the tile fall to attain a velocity of 0.716 sec? s = ut + ?*a*t? s = 0 + ?*9.8*0.716? s = 2.5m The tile fell from height 2.5m above the top of the window.
