A roofing tile falls from rest off the roof of a building. An observer from across the street notices that it takes 0.35 s for the tile to pass between two windowsills that are 2.45 m apart. How far is the sill of the upper window from the roof of the building?
Let H be the distance of the sill (base)of the upper window from the roof. The tile passes it in time
T = sqrt (2H/g)
You also know that
T + 0.35 = sqrt [2(H + 2.45)/g]
Solve those two simulataneous equations for H. T is easily eliminated.
0.35 = sqrt[2(H + 2.45)/g] - sqrt(2H/g)
Now solve for H
To find the distance between the sill of the upper window and the roof, we can use the equation of motion for a falling object:
d = (1/2) * g * t^2
Where:
- d is the vertical distance traveled by the tile
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time taken for the tile to fall (0.35 s)
However, we first need to determine the vertical distance traveled by the tile during the 0.35 seconds it takes to pass between the two windowsills.
To calculate this, we can use the formula for average velocity:
v_avg = d / t
Rearranging the equation, we can solve for d:
d = v_avg * t
Now, since the tile falls from rest, its initial velocity is 0 m/s. Therefore, the average velocity for the entire distance between the windowsills is equal to the final velocity at the lower windowsill.
Next, we need to calculate the final velocity at the lower windowsill. We can use the equation of motion:
v = u + g * t
Where:
- v is the final velocity
- u is the initial velocity (0 m/s)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time taken for the tile to fall (0.35 s)
Plugging in the values:
v = 0 + 9.8 * 0.35
v = 3.43 m/s
Now we can substitute this value of average velocity (v_avg) into the equation for d:
d = v_avg * t
d = 3.43 * 0.35
d = 1.2005 m
Therefore, the vertical distance between the sill of the upper window and the roof is approximately 1.2005 meters.