A ball falls from a height ‘h’ and collides with the ground. Each time it collides, its velocity is halved. Which of the following graphs are correct? Here t, v and y represent time, velocity and height above ground respectively. (vertically upwards direction is taken as positive and ground as y = 0)

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the motion of the ball as it falls and bounces back up, considering the conditions given in the question.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The ball falls from a height \( h \) and collides with the ground.
- Each time it collides, its velocity is halved.
- The downward direction is taken as negative, and the upward direction is positive.

2. **Initial Conditions**:
- When the ball is dropped from height \( h \), it starts with an initial velocity \( v_0 = 0 \) (just before it starts falling).
- As it falls, it accelerates due to gravity, reaching a maximum downward velocity just before it hits the ground.

3. **Velocity After Each Collision**:
- Let’s denote the velocity just before the first collision as \( v_1 \).
- After the first collision, the velocity becomes \( v_2 = \frac{v_1}{2} \) (half of \( v_1 \)).
- After the second collision, the velocity becomes \( v_3 = \frac{v_2}{2} = \frac{v_1}{4} \).
- This pattern continues, where after the \( n^{th} \) collision, the velocity is \( v_n = \frac{v_1}{2^{n-1}} \).

4. **Height After Each Collision**:
- The height reached after each bounce will also decrease because the velocity is halved each time.
- The maximum height after the first bounce can be calculated using the kinematic equation:
\[ h_1 = \frac{v_1^2}{2g} \]
- After the second bounce, the height reached will be:
\[ h_2 = \frac{v_2^2}{2g} = \frac{(\frac{v_1}{2})^2}{2g} = \frac{v_1^2}{8g} \]
- Continuing this, after the \( n^{th} \) bounce, the height will be:
\[ h_n = \frac{v_1^2}{2g \cdot 2^{n-1}} = \frac{v_1^2}{2g \cdot 2^{n-1}} \]

5. **Graph Analysis**:
- For the **velocity graph (v vs. t)**: The velocity will be maximum just before the first collision, then it will decrease to half after the first bounce, and continue to decrease after each subsequent bounce. The graph will show a series of peaks decreasing over time.
- For the **height graph (y vs. t)**: The height will start from \( h \) and decrease to 0 when the ball hits the ground. After each bounce, the height will reach a lower maximum height, creating a series of decreasing peaks on the graph.

6. **Conclusion**:
- The correct graphs will show a decreasing velocity after each collision and a decreasing height after each bounce.