A ball is dropped on the ground from a height of `1m`. The coefficient of restitution is `0.6`. The height to which the ball will rebound is

Options

A. 0.6 m
B. 0.4 m
C. 0.36 m
D. 0.16 m

Answer

To solve the problem of finding the height to which the ball will rebound after being dropped from a height of 1 meter with a coefficient of restitution of 0.6, we can follow these steps:

### Step 1: Calculate the velocity just before impact
When the ball is dropped from a height of 1 meter, 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:
- \( v \) = final velocity just before impact
- \( u \) = initial velocity (0 m/s, since it is dropped)
- \( a \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))
- \( s \) = height (1 m)

Substituting the values:
\[ v^2 = 0 + 2 \times 10 \times 1 \]
\[ v^2 = 20 \]
\[ v = \sqrt{20} = 2\sqrt{5} \, \text{m/s} \]

### Step 2: Calculate the velocity after the rebound
The coefficient of restitution (e) is defined as the ratio of the velocity of separation to the velocity of approach.

Given:
- Coefficient of restitution \( e = 0.6 \)
- Velocity of approach \( u = 2\sqrt{5} \)

Using the formula:
\[ v = e \cdot u \]
Substituting the values:
\[ v = 0.6 \cdot 2\sqrt{5} = 1.2\sqrt{5} \, \text{m/s} \]

### Step 3: Calculate the height to which the ball rebounds
Now, we will use the velocity after the rebound to find the maximum height (h) it reaches.

Using the same equation of motion:
\[ v^2 = u^2 + 2as \]
Here, at the maximum height, the final velocity \( v = 0 \), the initial velocity \( u = 1.2\sqrt{5} \), and acceleration \( a = -10 \, \text{m/s}^2 \).

Rearranging the equation gives:
\[ 0 = (1.2\sqrt{5})^2 + 2(-10)h \]
\[ 0 = 1.44 \cdot 5 - 20h \]
\[ 20h = 7.2 \]
\[ h = \frac{7.2}{20} = 0.36 \, \text{m} \]

### Final Answer
The height to which the ball will rebound is **0.36 meters**.

---