A roof tile falls from rest from the top of a building. An observer inside the building notices that it takes 0.12 s for the tile to pass her window, which has a height of 2.2 m. How far above the top of this window is the roof?
m
if it takes t seconds to reach the top of the window, it takes t+.12 seconds to reach the bottom. In that interval, it falls 2.2m
4.9(t+.12)^2 - 4.9t^2 = 2.2
t = 1.81
In the 1.81 seconds, it fell
4.9*1.81^2 = 16.05
so, the roof is 16.05m above the top of the window.
To determine the distance above the top of the window that the roof is, we can use the kinematic equation for motion under constant acceleration:
s = ut + (1/2)at^2
where:
s is the distance covered,
u is the initial velocity (which is 0 as the tile was at rest),
t is the time taken, and
a is the acceleration.
In this case, we need to find the distance above the top of the window, so the initial distance is the height of the window. Let's assume the tile falls vertically downward, so the acceleration due to gravity will be -9.8 m/s^2 (negative because it's acting in the opposite direction of the positive coordinate system). The equation becomes:
s = 2.2 m + (1/2)(-9.8 m/s^2)t^2
Now we need to solve for t, the time taken for the tile to pass the window. We have the time given in the problem as 0.12 s. Plugging in this value and rearranging the equation:
(1/2)(-9.8 m/s^2)t^2 = s - 2.2 m
-4.9 m/s^2 t^2 = s - 2.2 m
t^2 = (s - 2.2 m) / (-4.9 m/s^2)
t^2 = (2.2 m - s) / 4.9 m/s^2
Now that we have the equation for t^2, we can substitute it into the equation derived from the given time:
0.12 s = t
0.12 s = sqrt((2.2 m - s) / 4.9 m/s^2)
Squaring both sides:
0.0144 s^2 = (2.2 m - s) / 4.9 m/s^2
Multiplying both sides by 4.9 m/s^2:
0.07056 m/s^2 = 2.2 m - s
Rearranging the equation to solve for s:
s = 2.2 m - 0.07056 m/s^2
Calculating the value:
s = 2.12944 m
Therefore, the roof is approximately 2.12944 meters above the top of the window.