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

When body moves upward, v =0,s =h
v= u+at
`o= u-gt implies t=(u)/(g)` and `h=(u^(2))/(2g)`
`s= ut+ (1)/(2)at^(2), h= (1)/(2)gt^(2)`
`(u^(2))/(2g)= (gt.^(2))/(2), t.= (u)/(g)`
Total time, `t=t+t.= (2u)/(g)`