If an object is thrown vertically up with initial speed u from the ground, then the time taken by the object to return back to ground is

Options

A. `(u^(2))/(2g)`
B. `(u^(2))/(g)`
C. `(u)/(2g)`
D. `(2u)/(g)`

Answer

D

Explanation

`s = ut - (1)/(2) g t^(2)`
When returns to Ground
`s = h_("max")- h_("max") = 0 `
`0 = ut - (1)/(2) g t^(2)`
` ut = (1)/(2) g t^(2)`
`t = (2u)/(g)`