A balloon is released from the ground and rises vertically up with an acceleration of 1.4 ms^-2. A ball is released from the balloon, 20 seconds after the balloon has left the ground. The maximum height reached by the ball from the ground is:

For balloon, u=0

a=1.4m/s²

Let the balloon ascend for t = 20s

Then height ascended = h₁

We know that

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

Here, $h_1=\frac{1}{2}\times \frac{14}{10}\times 20\times 20$

${\text{h}}_1=14\text{m}\times 20\Rightarrow \text{h}=280\text{m}$

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

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

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

The height of ball from the balloon,

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

Therefore, height reached by the ball from the ground is:

H=h1+h2=280+40=320m