Aroof tile falls from rest from the top of a building. an observer inside the building notices that it takes 0.25 s for the tile to pass her window, which has a height of 1.7 m. how far above the top of this window is the roof?

1.586m

Explanation:

let 'u' be the velocity of the roof tile when it reaches the window

now using the equation of motion

where s= distance travelled

u=velocity

a=acceleration of the object

t=time taken to travel the distance 's'

given:

s=1.7m (distance covered to pass the window)

t=0.25s (Time taken to pass the window)

a=g=9.8m/s^2 (since the roof tile is moving under the action of gravity)

thus, substituting the values in the above equation we get

u=5.575m/s

This is the velocity when the tile  touches the window top.

Let's take this in second scenario as the tile's final velocity(v).

Now we have another equation of motion as

initial speed (when starts to fall) will be zero.

So the distance travelled (h) i.e the height from which the tile falls from the top of the window is given by,

substituting the values in the above equation, we get

h=1.586m

Hence, the window roof is 1.586m far away from the roof

b.

explanation:

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