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(@H sin theta))/(g)`

Answer

B

Explanation

`H = (u^(2)sin^(2)theta)/(2g)` and `T = (2u sin theta)/(g) rArr T^(2) = (4u^(2)sin^(2)theta)/(g^(2))`
`:. (T^(2))/(H) = (8)/(g) rArr T = sqrt((8H)/(g)) = 2 sqrt((2H)/(g))`