A ballon is ascending at `15ms^(-1)`. A bag containing food is dropped from it when it is at a height of 500 m from the ground. Calculate the time it takes to reach the ground. `(g=9.8ms^(-2))`

Explanation

`u=1ms^(-1)" "h=500ms`
Using the formula, `" "-h=(+v_(0))t+(1)/(2)(-g t)t^(2)`
`h=(-v_(0))t+(1)/(2) g t^(2)`
`"we get "500=-15t+(1)/(2)xx9.8t^(2)`
`9.8t^(2)-30t-1000=0`
`"or "4.9t^(2)-15t-500=0`

`therefore" "t=(15pm sqrt((15)^(2)-4xx4.9xx(-500)))/(2xx4.9)`
`"i.e. "t=(15pm 100.1)/(9.8)`
`therefore" "t= 11.74s`
The bag will take 11.7 s to reach the ground (air resistance and buoyancy are not taken into account).