Two balls are dropped from the top of a cliff at a time interval `Deltat = 2s`. The first ball hits the ground, rebounds elastically (reversing direction instantly without losing speed), and collides with the second ball at a height h = 55m above the ground. How high is the top of the cliff?

Options

A. 120 m
B. 140 m
C. 180 m
D. Data insufficient

Explanation

To solve the problem, we need to analyze the motion of both balls and use the principles of kinematics.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let the height of the cliff be \( H \).
- The first ball is dropped at \( t = 0 \) seconds and the second ball is dropped at \( t = 2 \) seconds.
- The first ball hits the ground, rebounds elastically, and collides with the second ball at a height \( h = 55 \) m above the ground.

2. **Calculate the time taken for the first ball to hit the ground**:
- The equation of motion for the first ball is given by:
\[ H = \frac{1}{2} g t_1^2 \]
- Here, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( t_1 \) is the time taken by the first ball to hit the ground.

3. **Determine the time of flight for the first ball**:
- Rearranging the equation gives:
\[ t_1 = \sqrt{\frac{2H}{g}} \]

4. **Calculate the time taken for the first ball to rebound and reach the height \( h \)**:
- After hitting the ground, the first ball will take the same time to rise back to height \( h \) as it took to fall. The time taken to rise to height \( h \) can be calculated using:
\[ h = \frac{1}{2} g t_{rebound}^2 \]
- Rearranging gives:
\[ t_{rebound} = \sqrt{\frac{2h}{g}} \]

5. **Total time for the first ball**:
- The total time from when the first ball is dropped to when it collides with the second ball is:
\[ t_{total} = t_1 + t_{rebound} \]

6. **Time for the second ball**:
- The second ball is dropped at \( t = 2 \) seconds, so the time it has been falling when they collide is:
\[ t_{second} = t_{total} - 2 \]

7. **Setting up the equations**:
- Since both balls are at the same height \( h \) when they collide, we can set up the equation for the second ball:
\[ h = \frac{1}{2} g t_{second}^2 \]

8. **Substituting values**:
- We can now substitute the expressions for \( t_1 \) and \( t_{rebound} \) into the equation and solve for \( H \).

9. **Final Calculation**:
- By substituting \( h = 55 \) m and solving the equations, we can find the height \( H \) of the cliff.