A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in $$t_1$$. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in $$t_2$$. Time required to reach the ground, if it is dropped from the top of the tower, is :

Let the tower height be $$h$$, initial speed be $$u$$, and $$g$$ be acceleration due to gravity.

Projected upward: Taking downward as positive:

$$h = -ut_1 + \frac{1}{2}gt_1^2$$ ... (1)

Projected downward:

$$h = ut_2 + \frac{1}{2}gt_2^2$$ ... (2)

Dropped (u = 0):

$$h = \frac{1}{2}gt^2$$ ... (3)

From (1): $$h = \frac{1}{2}gt_1^2 - ut_1$$
From (2): $$h = \frac{1}{2}gt_2^2 + ut_2$$

Adding (1) and (2): $$2h = \frac{1}{2}g(t_1^2 + t_2^2) + u(t_2 - t_1)$$

Subtracting (1) from (2): $$0 = \frac{1}{2}g(t_2^2 - t_1^2) + u(t_1 + t_2)$$

$$u = \frac{g(t_1^2 - t_2^2)}{2(t_1 + t_2)} = \frac{g(t_1 - t_2)}{2}$$

Substituting back into (2): $$h = \frac{g(t_1-t_2)}{2} \cdot t_2 + \frac{1}{2}gt_2^2 = \frac{g}{2}(t_1 t_2 - t_2^2 + t_2^2) = \frac{g t_1 t_2}{2}$$

From (3): $$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot \frac{g t_1 t_2}{2}}{g}} = \sqrt{t_1 t_2}$$.

The correct answer is Option A: $$\sqrt{t_1 t_2}$$.