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

Options

A. `(gsqrt(t_1t_(2)))/2`
B. `(g(t_1t_(2)))/2`
C. `gsqrt(t_1t_(2))`
D. `("g"t_1t_2)/((t_1+t_(2)))`

Answer

B

Explanation

Let u be initial velocity of vertical projection and t be the time taken by the body to reach a height h from ground .
`:.h=ut+1/2(-g)t^(2)" or ""gt"^(2) -2ut+2h=0`
`:. t = (2u pmsqrt(4u^(2) -4g xx2h))/(2g) =(upmsqrt(u^(2)-2gh))/g`
It means t has two values i.e.,
`t_(1) =(u+sqrt((u^(2)-2gh)))/g and t_2 =(u-sqrt(u^2-2gh))/g`
`:. t_(1)+t_(2)=(2u)/g" or " u = (g(t_(1)+t_(2))/2`