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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A ball is dropped from the roof of a building . An observer notes that the ball takes $0.1$ s to cross over a window 1 m in height . After crossing the window , the ball takes another 1.00 s to come to the bottom of the building . What is height of the building and how high is the window. Take $g=10 ms^{-2}$

$\begin{array}{1 1}(A)\;30m\\(B)\;21\;m \\(C)\;9 m \\(D)\;67 m \end{array} $

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Let the ball take t seconds to fall up to the bottom of the window.
It takes $(t- 0.1) $ seconds to fall up the top of window.
since the height (or length ) of window is 1 m , we have
$1= \large\frac{1}{2}$$gt^2-\large\frac{1}{2} $$g (t-0.1)^2 $
Solving, $t= 1.05 s$
The ball takes another 1 s to come upto the bottom of the building .
Therefore, the total time for t taken by the ball of fall from the roof of the building upto the bottom is $(1.05+ 1.00) =2.05$ .
Hence the height h of the building is given by
$h= \large\frac{1}{2} $$gt^2 =\large\frac{1}{2} $$ \times 10 \times 1.05^2 $
$\quad= 21.0 m$
Hence B is the correct answer.
answered Jul 25, 2014 by meena.p

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