A stone is dropped into water from 80 m above and a second later, another stone is thrown vertically down. Both stones strike the water simultaneously. What is the initial speed of the second stone? Given `g=9.8ms^(-2)`.

Explanation

Given `v_(0)=0, g=9.8ms^(-2), h= 80m, t = ?`
We know that
`x=v_(0)t+1//2g t^(2)`
`"i.e. "80=1//2xx9.8t^(2)" "therefore t^(2)=(160)/(9.8)`
`"or "t=sqrt((160)/(9.8))=4.04s" "t_(2)=4.04-1=3.04s`
For the speed stone, `80=v_(0)(3.04)+1//2xx9.8xx(3.04)^(2)`
`"i.e. "80-45.28=3.04u`
`therefore" "v_(0)=11.42ms^(-1)`