Height traveled by ball (with balloon) in 2 sec
h1=1/2at2=1/2×5×22= 10 m
Velocity of the balloon after 2 sec
v = at = 5×2 = 10 m/s
Now if the ball is released from the balloon then it acquire same velocity in upward direction.
Let it move up to maximum height h2
v2=u2−2gh2
=> 0=(10)2−2×(10)×h2
∴ h2 = 5m
Greatest height above the ground reached by the ball = h1+h2 = 10+5 = 15 m