Analysis of situation: t = 0 is the time when the balloon started rising up. At t = 0, when the packet is dropped, the balloon is moving up with velocity v = 0 + at0 = at0. Hence, initial velocity of the packet will be v0=at0 (upward). As the balloon has started rising upwards with constant acceleration a, so after t0 seconds, its height from the ground is y0=12at20.
For packet: s = ut -12gt2
⇒ −12at20 = at0t−12gt2 ⇒ gt2−2at0t−at20=0
Solving the quadratic equation, we get t = at0g[1+√1+ga]