A particle is projected at angle `theta` from horizontal. The product of the two possible times it take to pass over the height `h (h lt H_("max"))` is

Options

A. `(2u)/(g)`
B. `(2h)/(g)`
C. `(2u sin theta)/(g)`
D. None

Answer

B

Explanation

`(H - h) = (1)/(2) g t^(2)`
`t_(1) = sqrt(2 ((H - h))/(g))`
`t_(2) = sqrt((2H)/(g))`
`T_(1) T_(2) = (sqrt((2H)/(g)) - sqrt((2(H - h))/(g))) (sqrt((2H)/(g)) + sqrt(2(H- h))/(g))`
`= (2h)/(g)`