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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 tower of height h.The total distance covered by it in the last second of its motion is equal to the distance covered by it in first three seconds.The value of h is $(g=10 m/s^2)$

\[(a)\;125 \;m \quad (b)\;200\; m \quad (c)\;100\; m \quad (d)\;80\; m\]

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1 Answer

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The distance covered in first 3 second
$s=\large\frac{1}{2} $$gt^2$
$\quad=\large\frac{1}{2} $$ \times 10 \times 3^2$
$ \quad=45$
If the ball takes 'n' second to fall to ground . The distance covered in $n^{th}$ second
$s_n=u+\large\frac{g}{2}$$(2n-1)$
$\quad=0+\large\frac{10}{2}$$(2n-1)$
$\quad=10n-5$
Therefore $45=10n-5$
$n=5$
Therefore $ h=\large\frac{1}{2} $$gt^2$
$\qquad=\large\frac{1}{2} $$\times 10 \times 25$
$\qquad=125 \;m$
Hence a is the correct answer.
answered Jul 1, 2013 by meena.p
edited May 24, 2014 by lmohan717
 

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