Let A, B and C be points in a vertical line at height h, `(4h)/(5) and (h)/(5) ` and from the ground. A particle released from rest from the point A reaches the ground in time T. The time taken by the particle to go from B to C is ` T_(BC) ` . Then ` (T)/(T_(BC)) ^(2) = `

Options

A. `(gT^(2))/(6)`
B. `(gT^(2))/(12)`
C. `(gT^(2))/(18)`
D. `(gT^(2))/(24)`

Answer

D

Explanation

As, `T = (2u_(y))/(g) :. u_(y) = (gT)/(2)`
`rArr h_(A) = u_(y)t_(A) - (1)/(2) g t_(A)^(2) = u_(y) ((2u_(y))/(3g)) - (1)/(2)g ((2u_(y))/(3g))^(2)`
`= (4)/(9) (u_(y)^(2))/(g) = ((4)/(9g)) ((gT)/(2))^(2) = (gT^(2))/(9) [ :' u_(y) = (gT)/(2)]`
`rArr h_(B) = u_(y) ((5)/(6)xx (2u_(y))/(g)) - (1)/(2) xx g xx ((5)/(6) xx (2u_(y))/(g))^(2)`
`= (5)/(18) (u_(y)^(2))/(g) = (5)/(18g) ((gT)/(2))^(2) = (5)/(72) gT^(2)`
`:. h_(A) - h_(B) = (gT^(2))/(24)`