Height travelled by ball (with balloon) in 2 sec

\(h - \frac{1}{2} \times 0 \times t^{2} - \frac{1}{2} \times 4.9 \times 2^{2} - 9.8 m\)

Velocity of the balloon after 2 sec

\(v = a \tau = 4.9 \times 2 = 9.8 \text{ m/s}\)

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

Let it move up to maximum height \(\mathrm{H}_2\)

v^2 = u^2 + 2gh_2 ⇒ ⇒ \(0 = (9.8)^2 - 2 \times (9.8) \times h_i\) \hbar : =4.9 m

Greatest height above the ground reached by the ball \(g + h_2 \quad 9.8 + 4.9 \quad 14.7\,m\)