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Q.

Two particles A and B are placed as shown in figure. The particle A, on the top of a tower, is projected horizontally with a velocity u and the particle B is projected along the surface towards the tower, simultaneously. If both the particles meet each other. Then the speed of projection of particle B is [Ignore any friction]

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a

$d \sqrt{\frac{g}{2 H}} - u$

b

$d \sqrt{\frac{g}{2 H}}$

c

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d

$u$

answer is A.

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Detailed Solution

Time taken by both = $\sqrt{\frac{2 H}{g}}$
$R a n g e o f' A' = u \sqrt{\frac{2 H}{g}}$
$B t r a v e l s d = u \sqrt{\frac{2 H}{g}}$
$V e l o c i t y o f B = \frac{d - u \sqrt{\frac{2 H}{g}}}{\sqrt{\frac{2 H}{g}}}$
$\Rightarrow So speed of particle B = d \sqrt{\frac{g}{2 H}} - u$

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