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

Explanation

To solve the problem step by step, we need to determine the height to which the ball rebounds after being dropped from a height of 1 meter, given the coefficient of restitution (E) is 0.6.

### Step 1: Calculate the initial velocity just before impact
When the ball is dropped from a height \( H = 1 \, \text{m} \), it will fall freely under the influence of gravity. The velocity of the ball just before it hits the ground can be calculated using the formula:
\[ v = \sqrt{2gH} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

Substituting the values:
\[ v = \sqrt{2 \times 9.81 \times 1} = \sqrt{19.62} \approx 4.43 \, \text{m/s} \]

### Step 2: Calculate the velocity after the rebound
The coefficient of restitution (E) relates the velocities before and after the impact. It is defined as:
\[ E = \frac{\text{Velocity after impact}}{\text{Velocity before impact}} \]
Let \( v' \) be the velocity after the impact. Then:
\[ v' = E \times v \]
Substituting the values:
\[ v' = 0.6 \times 4.43 \approx 2.66 \, \text{m/s} \]

### Step 3: Calculate the maximum height after the rebound
To find the maximum height \( H' \) the ball reaches after rebounding, we can use the formula:
\[ H' = \frac{(v')^2}{2g} \]
Substituting the value of \( v' \):
\[ H' = \frac{(2.66)^2}{2 \times 9.81} = \frac{7.0756}{19.62} \approx 0.36 \, \text{m} \]

### Conclusion
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, is approximately **0.36 meters**.

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