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The windows on floors are separated by the same vertical distance. A brick is dropped...
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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,
where, L - distance to the first window from ground
h - verticle distance between windows
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,
Posted by malagala on May 9, 2012 at 6:29 PM (Answer #1)
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