A stone is thrown at an angle `theta` to the horizontal reaches a maximum height H. Then the time of flight of stone will be:

Options

A. `sqrt((2H)/g)`
B. `2sqrt((2H)/g)`
C. `(2sqrt(2H sin theta))/g`
D. `sqrt(2H sin theta)/g`

Answer

B

Explanation

`H = (u^(2) sin^(2)theta)/(2g)`
`T = (2usin theta)/g rArr T^(2) = (4u^(2) sin^(2)theta)/(g^(2)) = 4/g (u^(2) sin^(2)theta)(2g) xx 2`
`rArr T = sqrt((8H)/g) = 2sqrt((2H)/g)`