Two particles A and B are placed as shown in figure the particle A on the top of tower

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Two particles A and B are placed as shown in figure. The particle A on the top of a tower of height H is projected horizontally with a velocity u and the particle B is projected along the horizontal surface towards the foot of the tower, simultaneously. When particle A reaches at ground , it simultaneously hits particle B. Then the speed of projected particle B is:(neglect any type of friction)

• A

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

• B

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

• C

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

• D

  u

time $$=\dfrac{d}{v+u}$$

time $$=\sqrt{\dfrac{2H}{g}}$$

$$\dfrac{2H}{g}=\dfrac{d^2}{(v+u)^2}$$

$$(v+u)^2=\dfrac{d^2g}{2H}$$

$$v=\sqrt{\dfrac{d^2g}{2H}}-u$$

$$v=d\sqrt{\dfrac{g}{2H}}-u$$

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