A ball is dropped on a horizontal surface from height h. If it rebounds upto height `(h)/(2)` after first collision then coefficient of restitution between ball and surface is

Options

A. `(1)/(sqrt2)`
B. `(1)/(2)`
C. `(1)/(4)`
D. `(1)/(2sqrt2)`

Answer

To find the coefficient of restitution (e) between the ball and the surface after it rebounds from a height of \( \frac{h}{2} \), we can follow these steps:

### Step 1: Determine the velocity of approach (v_a)
When the ball is dropped from a height \( h \), we can use the equation of motion to find the velocity just before it hits the ground.

Using the equation:
\[ v^2 = u^2 + 2as \]
where:
- \( u = 0 \) (initial velocity, since the ball is dropped),
- \( a = g \) (acceleration due to gravity),
- \( s = h \) (distance fallen).

Substituting the values:
\[ v_a^2 = 0 + 2gh \]
Thus,
\[ v_a = \sqrt{2gh} \]

### Step 2: Determine the velocity of separation (v_s)
After the ball rebounds to a height of \( \frac{h}{2} \), we can find the velocity just after it leaves the ground using the same equation of motion.

Using the equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 0 \) (final velocity at the maximum height),
- \( u = v_s \) (initial velocity just after the bounce),
- \( a = -g \) (acceleration due to gravity, acting downwards),
- \( s = \frac{h}{2} \) (height reached after bounce).

Substituting the values:
\[ 0 = v_s^2 - 2g \left(\frac{h}{2}\right) \]
This simplifies to:
\[ v_s^2 = gh \]
Thus,
\[ v_s = \sqrt{gh} \]

### Step 3: Calculate the coefficient of restitution (e)
The coefficient of restitution is defined as the ratio of the velocity of separation to the velocity of approach:
\[ e = \frac{v_s}{v_a} \]
Substituting the values we found:
\[ e = \frac{\sqrt{gh}}{\sqrt{2gh}} \]
This simplifies to:
\[ e = \frac{1}{\sqrt{2}} \]

### Conclusion
Thus, the coefficient of restitution between the ball and the surface is:
\[ \boxed{\frac{1}{\sqrt{2}}} \]