A ball is projected from the bottom of a tower and is found to go above the tower and is caught by the thrower at the bottom of the tower after a time interval `t_1`. An observer at the top of the tower see the same ball go up above him and then come back at this level in a time interval `t_2` . The height of the tower is

Options

A. `1/2 g t_1 t_2`
B. `(g t_1t_2)/(8)`
C. `g/8 (t_1^2 - t_2^2)`
D. `g/2(t_1 - t_2)^2`

Answer

To find the height of the tower based on the given information, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion of the Ball:**
- The ball is projected upwards from the bottom of the tower and reaches a maximum height before falling back down.
- The time taken for the ball to go up and be caught by the thrower at the bottom is \( t_1 \).
- An observer at the top of the tower sees the ball go up and then come back to the same level in a time interval \( t_2 \).

2. **Time to Reach Maximum Height:**
- The time taken to reach the maximum height from the bottom of the tower is \( \frac{t_1}{2} \).
- Using the kinematic equation \( v = u + at \), where \( v = 0 \) (at maximum height), \( u \) is the initial velocity, \( a = -g \) (acceleration due to gravity), and \( t = \frac{t_1}{2} \):
\[ 0 = u - g \left(\frac{t_1}{2}\right) \]
- Rearranging gives:
\[ u = g \left(\frac{t_1}{2}\right) \]

3. **Time to Reach Maximum Height from the Top:**
- The time taken to reach maximum height from the top of the tower is \( \frac{t_2}{2} \).
- Using the same kinematic equation for the motion from the top to the maximum height:
\[ 0 = u' - g \left(\frac{t_2}{2}\right) \]
- Rearranging gives:
\[ u' = g \left(\frac{t_2}{2}\right) \]

4. **Total Time of Flight:**
- The total time of flight for the ball from the bottom to the top and back down is \( t_1 \).
- The time taken to go from the bottom to the top of the tower can be expressed as:
\[ t' = \frac{t_1}{2} + \frac{t_2}{2} \]
- Therefore, the time taken to reach the top from the bottom of the tower is:
\[ t' = \frac{t_1}{2} - \frac{t_2}{2} \]

5. **Calculating the Height of the Tower:**
- Using the kinematic equation for displacement \( s = ut + \frac{1}{2}at^2 \):
\[ h = u t' + \frac{1}{2}(-g)(t')^2 \]
- Substituting \( u \) and \( t' \):
\[ h = g\left(\frac{t_1}{2}\right)\left(\frac{t_1}{2} - \frac{t_2}{2}\right) - \frac{1}{2}g\left(\frac{t_1}{2} - \frac{t_2}{2}\right)^2 \]

6. **Final Expression for Height:**
- After simplifying the above expression, we arrive at:
\[ h = \frac{g}{8}(t_1^2 - t_2^2) \]

### Final Answer:
The height of the tower is given by:
\[ h = \frac{g}{8}(t_1^2 - t_2^2) \]