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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A stone is dropped from top of a tower when it crosses a point 5 m below the top, another stone is let fall from a point 25 m below the top. Both stones reach the bottom of tower simultaneously. The height of tower is $(g=10 m/s^2)$

\[(a)\;90\; m \quad (b)\;45 \;m \quad (c)\;35 \;m \quad(d)\;25\; m \]

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Let height of tower be 'h'
Time taken for 1st stone to reach ground is $=\sqrt {\large\frac{2h}{g}}$
$\bigg[s=ut+\large\frac{1}{2}$$gt^2$$; h=\large\frac{1}{2} $$gt^2\bigg]$
Time taken for the first stone to be travel 5 m from top $=\sqrt {\large\frac{2 \times 5}{g}}$
Time taken for 2nd stone to reach ground from a point $(h-25)m$ above the ground is $\sqrt {\large\frac{2(h-25)}{g}}$
Therefore $ \sqrt {\large\frac{2h}{g}}-\sqrt {\large\frac{2 \times 5}{g}}=\sqrt {\large\frac{2(h-25)}{g}}$
$\sqrt {2h}-\sqrt {10}=\sqrt {2(h-25)}$
Solving we get $h=45 \;m$
Hence b is the correct answer.


answered Jul 2, 2013 by meena.p
edited Jan 26, 2014 by meena.p

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