એક બલૂન 4.9 m/s^2 ના પ્રવેગથી ઉપર તરફ ગતિ

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$