If a ball is thrown vertically upwards with a speed  ‘u’, the distance covered during the last ‘t’ seconds of its ascent is……..

We know that the time of ascent  T=ugTotal distance covered by the ball in ‘T’ secondsH=uT−12gT2=uug−12gug2=u2g−u22g=u22g………..(i)Similarly distance covered during  (T-t)sech=u[T−t)−12(T−t)2=uug−t−12gug−t2=u2g−ut−12gu2g2+t2−2utg=u2g−ut−u22g+12gt2−ut=u2g−ut−u22g−12gt2+ut=u22g−12gt2Distance covered during the last ‘t’sec of its ascent  =H-h=u22g−u22g−12gt2=12gt2