If a ball is thrown vertically upwards with speed $u$, the distance covered during the last $t$ seconds of its ascent is 

The distance covered by the ball during the last t seconds of its upward motion = Distance covered by it in first t seconds of its downward motion

From$h=ut+\frac{1}{2}g{t}^2$ $h=\frac{1}{2}g{t}^2$         [As u = 0 for it downward motion]