A body is thrown vertically up to reach its maximum height in t seconds. The total time from the time of projection to reach a point at half of its maximum height while returning (in seconds) is

Options

A. `sqrt2t`
B. `(1+1/sqrt2)t`
C. `(3t)/2`
D. `t/sqrt2`

Answer

B

Explanation

Time for upward journey = t
Maximum height = h
If `t_1` is the time for falling through height h, then
`h=(1)/(2) "gt"_(1)^(2) " "( :. u =0 ) " "...(i)`
If `t_2` is the time for falling through height h , then
`h=(1)/(2) "gt"_(2)^(2) " "( :. u =0 ) " "...(ii)`
From equation (i) and (ii) , we get, `t_(2) =t_(1)/sqrt2`
Since time of rise = time of fall so `t =t_(1) :. t_(2) =t /sqrt2`
`:.` Total time required `=t+t_(2)+t/sqrt2=t(1+1/sqrt2)`