A balloon starts rising from the ground with an acceleration `2 ms^(-2)`. After `5 second , a stone is released from the ballooon. Find the time taken by the stone to reach the ground after its release and total height attained by ballooon when stone reaches the ground. Take `g==10 ms^(-2)`.

Explanation

Taking upward motion of balloon and stone,
`u=0, a=2 ms^(-2) ,t=5 s,
velocity attained by balloon after `5 s` is
``v=u+at =0 2+5 =10ms^(-1)` ltbtgt S =ut +1/2 at^(2) =0 + 1/2 2 xx 5^(2) =25 m.
Now the stone a height of `25 m` wigh inital velcity `10 ms^(-1)`. acting upwards.
Taking vertical downward motion of stone from balloon to ground we have
`u=-10 ms^(-1) , a= g =10 ms^(-1) , S=25 m`,
t=?`
S= ut + 1/2 at^(2)`
`25 =- 10 xx t 1/2 xx 10 xx t^(2) =- 10 t + 5 t^(2)`
or `t^(2) -2 t-5 =0`
or `t= (2+- swrt 4 +20)/2 =(2+- 2 sqrt6)/6 =1 =_ sqrt 6`
`-+- 2.449 =3.449 s or -1. 449s`
Since time (t) a positive quantty, so time taken by stone toreach the ground `=3.449 =3.45 s`
Total time for upward motion of balloon `T=5 +3.45 =845 s`
Total height attained by balloon in time `8.45 s`
`h=ut + 1/2 a T^(2) =0 + 1/2 xx 2 xx 98.45) ^(2)`
`=71.4 m`.