If a ball is thrown vertically upwards with speedu 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−12gT2  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(T−t)−12g(T−t)2So distance covered by ball in last t secondsh=H−x=[uT−12gT2]−[u(T−t)−12g(T−t)2]  =ut−gtT+12gt2=12gt2[∵T=u/g]