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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A stone is allowed to fall from the top of tower 100 metre high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25\;ms^{-1} $. The two stones will meets after

$\begin{array}{1 1} 4\;s \\ 0.4\;s \\ 0.04\;s \\.40\;s \end{array} $

Can you answer this question?
 
 

1 Answer

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Answer : 4s
$x= \large\frac{1}{2}$$gt^2,100-x=25t - \large\frac{1}{2}$$gt^2$
Adding $25 t= 100$ or $t= 4s$
answered Aug 19, 2014 by meena.p
 

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