An object thrown upwards with a velocity $u$ from the top of a tower of height $H$ reaches the ground in $t_1$ seconds. When thrown downwards with the same speed it reaches the ground in $t_2$ seconds. When dropped from the top of tower it reaches in $t_3$ seconds. Then $t_3$ is equal to (in terms of $t_1$ and $t_2$)

The correct option is B $\sqrt{t_1{t}_2}$
assuming downward direction as positive we get

$H=-u{t}_1+\frac{1}{2}a{t}_1^2$

and $H=u{t}_2+\frac{1}{2}a{t}_2^2$

multiplying 1 with $t_2$ and 2 with $t_1$ and then add both we get

$=>H(t_1+t_2)=\frac{1}{2}a{t}_1{t}_2(t_1+t_2)$

$=>2H=a{T}_1.t_2$

so when ball is simply released from top we get

$H=\frac{1}{2}a{t}_3^2$

using value of H we get

$=>t_3^2=t_1.t_2$

so best answer is option B.