A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after 1 t and 2 t seconds respectively, then

The correct option is: t = √(t₁t₂)

To find the relationship between the times $t$, $t_1$, and $t_2$, we analyze the equations of motion for the three scenarios under constant acceleration due to gravity $g$. Let the height of the building be $h$. We will take the downward direction as positive.

Case 1: The stone is dropped from the building
The stone is released from rest, so its initial velocity $u=0$.
Using the second equation of motion, $s=ut+\frac{1}{2}a{t}^2$:
$h=0+\frac{1}{2}g{t}^2$
$h=\frac{1}{2}g{t}^2$ — (Equation 1)

Case 2: A stone is thrown vertically upwards with velocity $u$
Since the stone is thrown upwards (opposite to the downward displacement), its initial velocity is $-u$.
Using the second equation of motion for time $t_1$:
$h=-u{t}_1+\frac{1}{2}g{t}_1^2$ — (Equation 2)

Case 3: A stone is thrown vertically downwards with velocity $u$
Since the stone is thrown downwards, its initial velocity is $+u$.
Using the second equation of motion for time $t_2$:
$h=u{t}_2+\frac{1}{2}g{t}_2^2$ — (Equation 3)

Now, we will eliminate the initial velocity $u$ from Equation 2 and Equation 3.
From Equation 2, we get:
$u{t}_1=\frac{1}{2}g{t}_1^2-h$
$u=\frac{1}{2}g{t}_1-\frac{h}{t_1}$ — (Equation 4)
From Equation 3, we get:
$u{t}_2=h-\frac{1}{2}g{t}_2^2$
$u=\frac{h}{t_2}-\frac{1}{2}g{t}_2$ — (Equation 5)

Equating the expressions for $u$ from Equation 4 and Equation 5:
$\frac{1}{2}g{t}_1-\frac{h}{t_1}=\frac{h}{t_2}-\frac{1}{2}g{t}_2$
Rearranging the terms to group the $g$ and $h$ terms respectively:
$\frac{1}{2}g{t}_1+\frac{1}{2}g{t}_2=\frac{h}{t_1}+\frac{h}{t_2}$
$\frac{1}{2}g(t_1+t_2)=h(\frac{1}{t_1}+\frac{1}{t_2})$
$\frac{1}{2}g(t_1+t_2)=h\left(\frac{t_1+t_2}{t_1{t}_2}\right)$

Since the time values are positive, $t_1+t_2\ne 0$. We can divide both sides by $(t_1+t_2)$:
$\frac{1}{2}g=\frac{h}{t_1{t}_2}$
$h=\frac{1}{2}g{t}_1{t}_2$ — (Equation 6)

Comparing Equation 1 ($h=\frac{1}{2}g{t}^2$) and Equation 6 ($h=\frac{1}{2}g{t}_1{t}_2$):
$\frac{1}{2}g{t}^2=\frac{1}{2}g{t}_1{t}_2$
Simplifying the equation gives:
$t^2=t_1{t}_2$
Taking the square root of both sides:
$t=\sqrt{t_1{t}_2}$