A body projected vertically upwords from the top of a tower reaches the ground in `t_(1)` second . If it projected vertically downwards from the some top with same velocity ,it reaches the ground in `t_(2)`seconds . If it is just dropped from the top it reaches the ground in t second .prove that `t=sqrt(t_(1)t_(2))`

Explanation

Taking downward direction as postive and applying equation of motion `s=ut+((1)/(2))at^(2)`, we get
`h=-ut_(1)+(1)/(2)gt_(1)^(2)`............(i)
and `h=ut_(2)+(1)/(2)gt_(2)^(2)` ..............(ii)
Mu ltiplying Eqn. (i) by `t_(20` and Eqn (ii) by `t_(1)`and adding the two, we get
`h(t_(1)+t_(2))=(1)/(2)gt_(1)t_(2)(t(1)+t_(2))`
or `h=(1)/(2)gt_(1)t_(2) ["as"t_(1)+t_(2)cancel=0]`
or `(1)/(2)g t^(2)=(1)/(2)gt_(1)t_(@) ["as"h=(1)/(2)g t^(2)]`
or `t=sqrt(t_(1)t_(2))`