The windows on floors are separated by the same

vertical distance. A brick is dropped from a window on each floor at the same time. The bricks should hit the ground at

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for this problem we can use the equation,

###### s = ut + (1/2)(a.t^2)

Here the starting velocity is zero. Hence the equation becomes

s = (1/2)(a.t^2)

as this is motion under gavity a = g in the downword direction

=> t = sqrt(2s/g)

As the windows are seperated by the same verticle distance, the distance that the bricks drop can be written as,

L,L+h,L+2h,L+3h,....,L+nh

where, L - distance to the first window from ground

h - verticle distance between windows

if L=h

s={h,2h,3h,...,nh}

then t1 = sqrt(2h/g) = time taken for the first brick to reach ground

Then time taken for the other bricks to reach the ground can be written as,

**t={t1,t1.sqrt(2),t1.sqrt(3),...,t1.sqrt(n)}**

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