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

The distance H covered by the ball in upward motion is given byH=uT−12gT2where T is the time taken by the ball.Further v = u-gT or 0=u-gTor T=ug...(2)Therefore, H=u×ug−12gug2.....(3)Before the last t second, the time taken by the ball will be (T - t) second. Let the distance be x, Thenx=u(T−t)+12g(T−t)2or x=12g(T−t)2=12gug−t22 (∵u=v=0)Required dista4ce = (H - x)∴(H−x)=u2g−12u2g−12gug−t2=ut−12gt2