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A body X is dropped from a vertical height of 200 m. Exactly at the same instant another body -Y is projected vertically upwards with a velocity of $100 ms^{-1}$. When and where the two bodies X and Y meet?

$\begin{array}{1 1}(A)\;19.6 \;m\\(B)\;20\;m \\(C)\;9 m \\(D)\;67\; m \end{array} $

1 Answer

Let the two bodies meet after t seconds. If the downwards distance covered by the body X is $h_1$ and the upward distance covered by the body Y is $h_2$ then the total height h of the building is given by ,
$h= h_1+h_2$
$\quad= \large\frac{1}{2}$$gt^2 + \bigg[ ut -\large\frac{1}{2} $$gt^2\bigg]$
$\quad= ut$
Where u is the velocity of projection of the body-Y .
Here $u= 100 ms6{-1} $ and $h= 200 m$ therefore $t= \large\frac{h}{u} =\frac{200}{100} $$=2s$
In $t= 2s$ the body X will cover a distance of $\large\frac{1}{2}$$ \times g \times 2^2$ ie. $19.6\;m$
Hence A is the correct answer.
answered Jul 25, 2014 by meena.p

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