A body is projected vertically upwards . The times corresponding to height h while ascending and while descending are `t_(1)andt_(2)` respectively . Then the velocity of projection is (g is acceleration due to gravity)

Options

A. `(gsqrtt_(1)t_(2))/(2)`
B. `(g(t_(1)+t_(2)))/(2)`
C. `gsqrt(t_(1)t_(2))`
D. `(g t_(1)t_(2))/(t_(1)+t_(2))`

Answer

B

Explanation

When the body is projected , it takes time `t_(1)` to cross h , then another `t_(1)` to go to `h_(max)` when v=0.
Then the initial velocity become u=0 and takes time t to reach the ground.
`0=u-g t`
`t=u//g`.
`t_(1)+t_(1)=t=u//g`.
`thereforet_(1)+t_(1)=u//g`
`t_(1)=(u//g)-t_(1)`,.
`thereforet_(1)+t_(1)=(u)/(g)+t_(1)`
`thereforet_(1)+t_(2)=(2u)//g`
`rArru=(g)/(2)(t_(1)+t_(2))`