posted by pavatharani on .
a stone is thrown vertically upward with an initial velocity v0. the distance travelled by it in time 1.5v0/g
In time Vo/g it teaches maximum height, H = Vo^2/(2g)
In an additional time 0.5 Vo/g, it will come part way back down. The distance it falls in that time is
(1/2)g*(0.5Vo/g)^2 = (1/8)(Vo^2/g) = H/4
The total distance travelled is (5/4)H and the total displacement is (3/4)H