You throw a small rock straight up from the edge of a highway bridge that crosses a river. the rock passes you on its way down, 8.00 s after it was thrown. what is the speed of the rock just before it reaches the water 21.0 m below the point where the rock left your hand? ignore air resistance

V^2=u^2 +2aS
U is found first by considering that first 8 secs and using v=u+at. {different v and u though}

V=-u+gt.
Magnitude of u = magnitude of v if there is no resistance ( because the conservation of energy says the k. E. must be the same when it passes you as when it left your hand) up is negative here, down is positive.
V+v=gt
2v= g x 8
V=4xg.= the initial velocity for the next calculation
V^2=(4g)^2+(2xgx21)

So v can be calculated.