A ball is released from the top of a multistory tower. The ball taked ` 1 s` to fall pasta floor of the tower `8 m` height of a floor some distance from the top of thetower. Find the velocities of the ball at the top and at the bottom of the window.

Answer

`v_(top) =3 m//s & v_(battom) =13 m//s`

Explanation

Let `v_(1)` and `v_(b)` are the velocities at top and bottom of window,
`s=(v_(1)+v_(b))/(2) t rArr 8=((v_(t)+v_(b)))/(2)xx1`
`v_(1)+v_(b)=16` (i)
using equation `v=u+at`
`v_(b)-v_(1)+(10)(1)`
`v_(b)-v_(t)=10`
solving equation (i) and (ii) we get (ii)
`v_(1)=3 m//s` and `v_(b)=13 m//s`.