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

Options

A. `t=t_(1)-t_(2)`
B. `t=(t_(1)+t_(2))/2`
C. `t=sqrt(t_(1)t_(2)`
D. `t=t_(1)^(2)t_(2)^(2)`

Explanation

To solve the problem step by step, we will analyze the motion of the stones dropped and thrown from a height \( h \).

### Step 1: Analyze the stone dropped from height \( h \)

When a stone is dropped from a height \( h \), its initial velocity \( u = 0 \). The equation of motion can be written as:

\[ h = \frac{1}{2} g t^2 \]

Where:
- \( g \) is the acceleration due to gravity,
- \( t \) is the time taken to reach the ground.

From this equation, we can express \( h \):

\[ h = \frac{1}{2} g t^2 \]

### Step 2: Analyze the stone thrown upwards

When a stone is thrown upwards with an initial velocity \( u \), it will take time \( t_1 \) to reach the ground. The equation of motion for this case is:

\[ -h = u t_1 - \frac{1}{2} g t_1^2 \]

Rearranging gives:

\[ -h = u t_1 - \frac{1}{2} g t_1^2 \]

### Step 3: Analyze the stone thrown downwards

For the stone thrown downwards with the same initial velocity \( u \), it will take time \( t_2 \) to reach the ground. The equation of motion is:

\[ -h = -u t_2 - \frac{1}{2} g t_2^2 \]

Rearranging gives:

\[ -h = -u t_2 - \frac{1}{2} g t_2^2 \]

### Step 4: Eliminate \( u \)

Now, we have two equations:

1. \( -h = u t_1 - \frac{1}{2} g t_1^2 \) (Equation 1)
2. \( -h = -u t_2 - \frac{1}{2} g t_2^2 \) (Equation 2)

We can multiply Equation 1 by \( t_2 \) and Equation 2 by \( t_1 \):

\[ -h t_2 = u t_1 t_2 - \frac{1}{2} g t_1^2 t_2 \quad (1') \]
\[ -h t_1 = -u t_1 t_2 - \frac{1}{2} g t_2^2 t_1 \quad (2') \]

### Step 5: Add the two equations

Adding (1') and (2') gives:

\[ -h t_2 - h t_1 = u t_1 t_2 - \frac{1}{2} g t_1^2 t_2 - u t_1 t_2 - \frac{1}{2} g t_2^2 t_1 \]

This simplifies to:

\[ -h (t_1 + t_2) = -\frac{1}{2} g (t_1^2 t_2 + t_2^2 t_1) \]

### Step 6: Solve for \( h \)

From the above equation, we can express \( h \):

\[ h = \frac{1}{2} g \frac{t_1 t_2 (t_1 + t_2)}{t_1 + t_2} \]

### Step 7: Relate \( t_1, t_2, \) and \( t \)

From the first equation \( h = \frac{1}{2} g t^2 \) and the derived equation for \( h \), we can equate them:

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

Thus, we find:

\[ t^2 = t_1 t_2 \]

Taking the square root gives:

\[ t = \sqrt{t_1 t_2} \]

### Final Result

The relationship between the times is:

\[ t = \sqrt{t_1 t_2} \]