A balloon reses rest on the ground with constant accleeration `1 m s^(-2)`. A stone is dropped when balloon has risen to a height of `39.2 m`. Find the time taken by the stone to teach the ground.

Answer

4

Explanation

`h(1)/(2) at^(2) rArr tsqrt((2h)/(a)) =sqrt((2xx40 xx 8)/(10)) =8 s`.
`Veolcity after `8` s, `v=0+1xx8=8ms^(-1)`
`40 =10t +(1)/(2)10t^(2)`
`10t^(2)-2xx10t-2xx40=0`
`r^(2)-2t-8=0`
`rt-4t+2t-8=0`
`t=4s`.