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

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

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

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

answer is A.

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Detailed Solution

Time taken by particles to collide $t = \sqrt{\frac{2 H}{g}}$

Then $u \sqrt{\frac{2 H}{g}} + V \sqrt{\frac{2 H}{g}} = d$

$\Rightarrow u + V = d \sqrt{\frac{g}{2 H}} or v = d \sqrt{\frac{g}{2 H}} - u$

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