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]