A balloon starts rising upwards with constant acceleration a and afrer time `t_(0)`, second, a packet is dropped from it which reaches the ground aftre t seconds of dropping . Derermine the value of `t`
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Explanation

Analysis of situation: `t=0` is the time when the balloon started rising up . At `t=t_(0)`, whe the packet is drpped, the balloon is moving up with velocity `v=0+at_(0)`. Hence, initial velocity fo the packet will be ` v_(0) =at_(0)` (upward). As the balloon has started rising upward with constant accelerartion a so after `t_(0)` second, its height form the ground is `y_(0) =(1)/(2) at_(0)^(2)`.
For packet : `s=ut -(1)/(2) g t^(2)`
`rArr -(1)/(2) at_(0)^(2) =at_(0)t-(1)/(2) g t^(2) rArr g t^(2) -2at_(0)t -at_(0)^(2) =0`
Solving the quadratic euation, we get `t=(at_(0))/(g) [1+sqrt(1+(g)/(a))]`.