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A particle is moving up with balloon wit...

A particle is moving up with balloon with constant accelration (g/8) which starts from rest from ground and at height H particle is droped from balloon. After this event, time for which particle will be in air is `sqrt((kH)/(g))`. Find the value of k.

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The correct Answer is:
4
`v^(2)=2xx(9)/(8)H`
`v=sqrt(9H)/(2)`
Now let time is T
`-H=sqrt(9HT)/(2)-(1)/(2)g t^(2)`
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