A ball thrown up vertically returns to the thrower after 8 second. Calculate
(i) velocity with which it was thrown
(ii) maximum height it acquired.
(iii) velocity with which it hit the ground. Given `g=9.8 m//s^(2)`.

Explanation

As time of ascent =time of descent`=(1)/(2)` total time.
`:. t=(8)/(2)=4s`
(i)`u=?,v=0,a=-g9.8m//s^(2)`.
From `v=u+at`
`0=u-9.8xx4` or `u=39.2m//s`
(ii)From `v^(2)-u^(2)=2a s`
`0-(39.2)^(2)=2(-9.8)h`
`h=(-(39.2)^(2))/(-2xx9.98)=78.4m`
(iii) velocity with which body hits the ground = velocity of projection of body `=39.2m//s`.