A balloon moves verticaIIy upwards with a uniform velocity `v_(0)`. A weight is tied to baIIoon with a rope. When the baIIoon attains a height `h_(0)`, the rope snaps. How much time will the weight take to reach the ground?

Explanation

Let the time taken by the weight after the rope snaps to reach the ground be t. From the question the weight ascends with the same velocity as that of the balloon, i.e, `v_(0)`. So when the rope snaps the initial velocity of the weight = -`v_(0)` [ the negative sign comes as the velocity of the weight is directed upwards just when the rope snaps]
`:.` For the free fall of the weight,
`h_(0)=-v_(0)t+(1)/(2)"gt"^(2)" ""or",(1)/(2)"gt"^(2)-v_(0)t-h_(0)=0`
`:." " t=(v_(0)pmsqrt(v_(0)^(2)-4xx(1)/(2)g(-h_(0))))/(2.(1)/(2)g)=(v_(0)pmsqrt(v_(0)^(2)+2gh_(0)))/(g)`
So, time taken by the weight to reach the ground is `=(v_(0)pmsqrt(v_(0)^(2)+2gh_(0)))/(g)" "[because t gt0].`