A balloon is released from the ground which rises vertically up with acceleration$1.4{\text{ms}}^{-2}$ . A ball is released from the balloon, 20 second after the balloon has left the ground. The maximum height reached by the ball from the ground is

For balloon u = 0

$a=1.4{\text{m/s}}^{\text{2}}$

Let the balloon ascended for t = 20s

Then height ascended = h

We get

$\text{s}=\text{ut}+\frac{1}{2}{\text{at}}^{\text{2}}$

$\text{s}=0\left(\text{t}\right)+\frac{1}{2}\left(1.4\right){\left(20\right)}^2$

$s=\frac{1}{2}\times \frac{14}{10}\times 20\times 20$

$\text{h}=14\text{m}\times 20\Rightarrow \text{h}=280\text{m}$

$\text{v}=\text{u}+\text{at}$                               

$\text{v}=\text{0+}\left(1.4\right)20$

$\text{v}=28{\text{ms}}^{-1}$

${\text{h}}_2=\frac{{\text{v}}^{\text{2}}}{\text{2g}}=\frac{28\times 28}{2\times 9.8}=\frac{28\times 28\times 10}{2\times 98}=40\text{m}$

$\text{H}={\text{h}}_1+{\text{h}}_2=280+40=320\text{m}$