Distance covered by ball thrown upwards with speed 'u' in last 't' seconds before it reaches max. height is .............

Distance covered by ball thrown upwards with speed 'u' in last 't' sec

1.	Distance covered by ball thrown upwards with speed 'u' in last 't' seconds before it reaches max. height is .............
Answer» ` (u + g t)t ` ut `1/2 g t^(2)` `ut -(1)/(2) g t^(2)` Solution :` (##KPK_AIO_PHY_XI_P1_C03_E06_001_S01.png" width="80%"> Suppose time taken by ball to reach max, height is T. `THEREFORE v = u - gT In v =0` Final velocity is zero at max. height `therefore u = gT` `therefore T = u/g ""…(1)` velocity gained by ball in `(T-t)s,` `v =u -g (T-t)` `=u -g ((u)/(g) -t) (because` From (1)) `=u - u + g t` `therefore v = g t""...(2)` Distance COVERED in last .t. SECONDS, `d = vt - 1/2 g t^(2)` `= (g t)t - 1/2 g t^(2) (because ` For (2)) `= g t ^(2) -1/2 g t^(2)` `therefore d = 1/2 g t^(2)`