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 height `h` above the ground as

Options

A.
B.
C.
D.

Explanation

To solve the problem of how the velocity \( v \) of a ball varies with the height \( h \) above the ground after being dropped from a height \( d \) and bouncing back to a height of \( \frac{d}{2} \), we can follow these steps:

### Step 1: Analyze the motion of the ball when dropped
When the ball is dropped from height \( d \), it starts from rest, so its initial velocity \( u = 0 \). As it falls, it accelerates due to gravity, \( g \).

Using the equation of motion:
\[ v^2 = u^2 + 2gh \]
where \( h \) is the height fallen (which will be \( d \) when it hits the ground), we can find the velocity just before it hits the ground.

Substituting \( u = 0 \) and \( h = d \):
\[ v^2 = 0 + 2gd \implies v = \sqrt{2gd} \]

### Step 2: Analyze the motion of the ball after bouncing
After hitting the ground, the ball bounces back to a height of \( \frac{d}{2} \). At the maximum height of the bounce, the final velocity \( v = 0 \) and the initial velocity \( u \) can be found using the same equation of motion.

Using the equation:
\[ v^2 = u^2 + 2gh \]
where \( h = \frac{d}{2} \), we set \( v = 0 \):
\[ 0 = u^2 - 2g\left(\frac{d}{2}\right) \implies u^2 = gd \implies u = \sqrt{gd} \]

### Step 3: Establish the relationship between velocity and height
Now we have two key points:
1. Just before hitting the ground (at height \( h = 0 \)), the velocity is \( v = \sqrt{2gd} \).
2. At the maximum height after bouncing (at height \( h = \frac{d}{2} \)), the velocity is \( v = 0 \).

### Step 4: Determine the graph of velocity vs. height
- As the ball falls from height \( d \) to height \( 0 \), the velocity increases from \( 0 \) to \( \sqrt{2gd} \).
- As the ball bounces back from height \( 0 \) to height \( \frac{d}{2} \), the velocity decreases from \( \sqrt{2gd} \) to \( 0 \).

### Step 5: Graphical representation
The graph of velocity \( v \) versus height \( h \) will show:
- A curve starting from \( (d, 0) \) and moving to \( (0, \sqrt{2gd}) \) as the ball falls.
- Then it will curve back from \( (0, \sqrt{2gd}) \) to \( (\frac{d}{2}, 0) \) as the ball bounces back.

### Conclusion
The velocity \( v \) varies with height \( h \) in a parabolic manner, reflecting the changes in kinetic and potential energy during the fall and bounce.