a
(a) Height travelled by ball (with balloon) in \(2\) sec

\({h_1} = \frac{1}{2}a\;{t^2} = \frac{1}{2} \times 4.9 \times {2^2} = 9.8\;m\)

Velocity of the balloon after 2 sec

\(v = a\;t = 4.9 \times 2 = 9.8\;m/s\)

Now if the ball is released from the balloon then it acquire same velocity in upward direction.

Let it move up to maximum height \({h_2}\)

\({v^2} = {u^2} - 2g{h_2}\) \(⇒\) \(0 = {(9.8)^2} - 2 \times (9.8) \times {h_2}\)\(\therefore \)\({h_2}=4.9\,m\)

Greatest height above the ground reached by the ball \( = {h_1} + {h_2} = 9.8 + 4.9 = 14.7\;m\)