To solve this problem, we'll use the equations of motion under constant acceleration due to gravity. Let's define:

$ h $: Height of the tower

$ u $: Initial speed of projection (same magnitude in both cases)

$ g $: Acceleration due to gravity (positive downward)

$ t_1 $: Time taken to reach the ground when projected upwards

$ t_2 $: Time taken to reach the ground when projected downwards

$ t $: Time taken to reach the ground when simply dropped

Case 1: Projectile Thrown Upwards

When the body is projected upwards from the top of the tower, its initial velocity is $ -u $ (since upward direction is negative), and it reaches the ground in time $ t_1 $. The equation of motion is:

$ h = -u t_1 + \frac{1}{2} g t_1^2 $

Case 2: Projectile Thrown Downwards

When the body is projected downwards from the top of the tower, its initial velocity is $ u $, and it reaches the ground in time $ t_2 $. The equation is:

$ h = u t_2 + \frac{1}{2} g t_2^2 $

Case 3: Body Dropped

When the body is simply dropped, its initial velocity is $ 0 $, and it reaches the ground in time $ t $:

$ h = \frac{1}{2} g t^2 $

Step 1: Equate the Heights

From cases 1 and 2, equate the expressions for $ h $:

$ -u t_1 + \frac{1}{2} g t_1^2 = u t_2 + \frac{1}{2} g t_2^2 $

Step 2: Solve for $ u $

Simplify the equation:

$ -u t_1 - u t_2 = \frac{1}{2} g t_2^2 - \frac{1}{2} g t_1^2 $

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

$ -u (t_1 + t_2) = \frac{1}{2} g (t_2 - t_1)(t_2 + t_1) $

Divide both sides by $ t_1 + t_2 $:

$ -u = \frac{1}{2} g (t_2 - t_1) $

$ u = \frac{1}{2} g (t_1 - t_2) $

Step 3: Express $ h $ in Terms of $ t_1 $ and $ t_2 $

Substitute $ u $ back into one of the equations for $ h $:

$ h = -\left( \frac{1}{2} g (t_1 - t_2) \right) t_1 + \frac{1}{2} g t_1^2 $

Simplify:

$ h = \frac{1}{2} g t_1 t_2 $

Step 4: Equate the Height for the Dropped Case

From the dropped case:

$ h = \frac{1}{2} g t^2 $

Set the two expressions for $ h $ equal to each other:

$ \frac{1}{2} g t^2 = \frac{1}{2} g t_1 t_2 $

$ t^2 = t_1 t_2 $

Step 5: Solve for $ t $

$ t = \sqrt{t_1 t_2} $

Conclusion:

The time required for the body to reach the ground when dropped is the geometric mean of $ t_1 $ and $ t_2 $:

Answer: Option B

$ t = \sqrt{t_1 \, t_2} $