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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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)}

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial