A ground to ground projectile is at point A at `t = (T)/(3)`, is at point B at `t = (5T)/(6)` and reaches the ground at `t = T`. The difference in heights between points A and B is

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)`