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Question

# A stone is dropped from a height h. Simultaneously, another stone is thrown up from the ground which reaches a height of 4h. The two stones cross each other after time

A
h2g
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B
h8g
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C
8hg
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D
2hg
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Solution

## The correct option is B √h8gLet the initial velocity of stone which is thrown from the ground u2. At maximum height 4h, Velocity v2=0Using v2=u2−2as ⇒0=u22−2g×4h ⇒u2=√8gh ...1Let h1 be the distance covered by the stone which is dropped from height h, Let h2 be the distance covered by the stone which is thrown up from the ground. Let at time t both stone crossed each other.Using s=ut+12at2, ⇒−h1=−12gt2 ⇒h1=12gt2 ....2⇒h2=u2t−12gt2 ....3Adding h1 and h2 gives the height h.∴ Adding equations 2 and 3,⇒h=12gt2+u2t−12gt2 ⇒h=u2tUsing the value of u2=√8gh, ⇒h=√8ght ⇒t=√h8g

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