A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air resistance, its velocity v varies with the heiht h above the ground as

Options

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B.
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D.

Explanation

To solve the problem, we need to analyze the motion of the ball as it falls and bounces back up. We will derive the relationship between the velocity \( v \) of the ball and its height \( h \) above the ground.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The ball is dropped from a height \( D \) (initial height).
- At the height \( D \), the initial velocity \( v_0 = 0 \).

2. **Determine Velocity Just Before Hitting the Ground**:
- As the ball falls, it accelerates due to gravity. We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
- Here, \( u = 0 \) (initial velocity), \( a = g \) (acceleration due to gravity), and \( s = D \) (distance fallen).
- Thus, the velocity just before hitting the ground (let's call it \( v_1 \)) is:
\[ v_1^2 = 0 + 2gD \implies v_1 = \sqrt{2gD} \]

3. **Determine Velocity After Bouncing**:
- After hitting the ground, the ball bounces back to a height of \( D/2 \).
- At the maximum height of \( D/2 \), the velocity \( v_2 = 0 \).
- We can again use the kinematic equation to find the velocity just after the bounce (let's call it \( v_2 \)):
\[ v^2 = u^2 + 2as \]
- Here, \( u = v_1 \) (velocity just before hitting the ground), \( a = -g \) (deceleration going up), and \( s = D/2 \) (height reached after bouncing).
- Thus:
\[ 0 = v_2^2 - 2g \left(\frac{D}{2}\right) \implies v_2^2 = gD \]

4. **Establish Relationship Between Velocity and Height**:
- From the above equations, we have:
- \( v_1^2 = 2gD \)
- \( v_2^2 = gD \)
- We can express the velocity \( v \) in terms of height \( h \):
\[ v^2 = 2gh \quad \text{(for height } h \text{ falling down)} \]
- This means that the velocity squared is proportional to the height above the ground.

5. **Graphical Representation**:
- The relationship \( v^2 \propto h \) indicates that if we plot \( v^2 \) against \( h \), we will get a straight line.
- As the ball falls from height \( D \) to the ground, \( v \) will be negative, and as it bounces back to height \( D/2 \), \( v \) will be positive.

### Final Conclusion:
The velocity \( v \) varies with height \( h \) above the ground according to the equation \( v^2 = 2gh \). This indicates a parabolic relationship when plotted.