Height travelled by ball (with balloon) in 2 sec

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

Velocity of the balloon after 2 sec

\(v \quad a \tau \quad 4.9 \times 2 \quad 9.8\,m/s\)

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

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

v^{2} = u^{2} + 2gh_{2} ⇒ ⇒ \(0 = (9.8)^2 - 2 \times (9.8) \times v_i\) \hbar : =4.9 m

Greatest height above the ground reached by the ball \(h_1 + h_2 \quad 9.8 + 4.9 \quad 14.7\,m\)