A balloon starts from rest and moves vertically upwards with a uniform acceleration  $\frac{g}{8}m{s}^{-2}$ A stone falls from the balloon after 8s from the start. Find the time taken by the stone to reach the ground  $(g=9.8m{s}^{-2})$

Step-1: To find the distance of the stone above the ground about which it begins to fall from the balloon. 
$S=ut+\frac{1}{2}a{t}^2$
Here, s=h, u=0, a=g/8
$h=\frac{1}{2}\left(\frac{g}{8}\right)8^2=4g$
Step-2: The velocity of the balloon at this height can be obtained from
v= u+at
$V=0+\left(\frac{g}{8}\right)8=g$
This becomes the intital velocity  $\left(u^{\text{'}}\right)$ of the stone as the stone falls from the balloon at the height h.  $\therefore u\text{'}=g$
Step-3: For the total motion of the stone  $h=\frac{1}{2}g{t}^2-u^{\text{'}}t$
Here, $h=4g,u\text{'}=g,t=$ time of travel of stone.
$\therefore -4g=gt-\frac{1}{2}g{t}^2$
$\therefore t^2-2t-8=0$
Solving for ‘t’ we get t = 4 and -2s. Ignoring negative Value of time, t = 4 s