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

Let T be the time of ascent and H be the total height.  Then $T=u/g$ 

And  $H=uT-\frac{1}{2}g{T}^2$  Let $(T-t)$ be the time taken by the ball to travel part of H. The distance covered in(T- t )sec  is  $x=u\left(T-t\right)-\frac{1}{2}g{\left(T-t\right)}^2$

So distance covered by ball in last $t$ seconds

$h=H-x=\left[uT-\frac{1}{2}g{T}^2\right]-\left[u\left(T-t\right)-\frac{1}{2}g{\left(T-t\right)}^2\right]$  $=ut-gtT+\frac{1}{2}g{t}^2=\frac{1}{2}g{t}^2\left[\because T=u/g\right]$