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

Options

A. ` (u + g t)t `
B. ut
C. `1/2 g t^(2)`
D. `ut -(1)/(2) g t^(2)`

Answer

C

Explanation

` (##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)`