12 seconds later the man sitting inside the balloon release a stone. Find the time taken by the stone to hit the ground.

$u=0 ,t=12 s ,a=0.2 m/s$ $s=ut+ \dfrac 12 at^2 \ =0+ \dfrac 12 ×0.2 ×12^2 \ =14.4 m$ ∴ stone will be released from a height of 14.4 m Initial velocity of stone = final velocity of stone after 12 sec Final velocity of stone after 12 sec, $v=u+at \ =0+0.2 ×12 \ =2.4 m/s$ i.e. Initial velocity of stone = 2.4 m/s $S=ut + \dfrac 12 at^2 \ -14.4=2.4 t + \dfrac 12 ×9.81 ×t^2 \ 4.905 t^2- 2.4 t-14.4=0 \ t=1.9754 s$

1.9754 sec time taken by the stone to hit the ground.

$s=ut+ \dfrac 12 at^2 \\ =0+ \dfrac 12 ×0.2 ×12^2 \\ =14.4 m$

∴ stone will be released from a height of 14.4 m

Initial velocity of stone = final velocity of stone after 12 sec

Final velocity of stone after 12 sec,

$v=u+at \\ =0+0.2 ×12 \\ =2.4 m/s$

i.e. Initial velocity of stone = 2.4 m/s

$S=ut + \dfrac 12 at^2 \\ -14.4=2.4 t + \dfrac 12 ×9.81 ×t^2 \\ 4.905 t^2- 2.4 t-14.4=0 \\ t=1.9754 s$

1.9754 sec time taken by the stone to hit the ground.