A ball is dropped from a height 100 m on the ground. If the coefficient of restitution is 0.2, the height to which the ball will go up after it rebounds for the `II^(nd)` time.

Options

A. 16 m
B. 1.6 cm
C. 1.6 m
D. 16 cm

Answer

To find the height to which the ball will rebound after the second bounce, we can follow these steps:

### Step 1: Understand the Coefficient of Restitution
The coefficient of restitution (e) is defined as the ratio of the velocity of separation to the velocity of approach. It quantifies how much energy of motion (kinetic energy) is retained after a collision.

### Step 2: Calculate the Initial Velocity Before Impact
When the ball is dropped from a height (h = 100 m), it will gain speed due to gravity just before it hits the ground. The velocity (u) just before impact can be calculated using the formula:
\[ u = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

Substituting the values:
\[ u = \sqrt{2 \times 9.81 \times 100} \]
\[ u = \sqrt{1962} \]
\[ u \approx 44.4 \, \text{m/s} \]

### Step 3: Calculate the Velocity After the First Rebound
Using the coefficient of restitution, the velocity after the first rebound (v1) is given by:
\[ v_1 = e \times u \]
Substituting the values:
\[ v_1 = 0.2 \times 44.4 \]
\[ v_1 \approx 8.88 \, \text{m/s} \]

### Step 4: Calculate the Height After the First Rebound
The height (h1) to which the ball rebounds after the first impact can be calculated using the formula:
\[ h_1 = \frac{v_1^2}{2g} \]
Substituting the values:
\[ h_1 = \frac{(8.88)^2}{2 \times 9.81} \]
\[ h_1 = \frac{78.9}{19.62} \]
\[ h_1 \approx 4.02 \, \text{m} \]

### Step 5: Calculate the Velocity After the Second Rebound
Using the coefficient of restitution again, the velocity after the second rebound (v2) is given by:
\[ v_2 = e \times v_1 \]
Substituting the values:
\[ v_2 = 0.2 \times 8.88 \]
\[ v_2 \approx 1.776 \, \text{m/s} \]

### Step 6: Calculate the Height After the Second Rebound
The height (h2) to which the ball rebounds after the second impact can be calculated using the formula:
\[ h_2 = \frac{v_2^2}{2g} \]
Substituting the values:
\[ h_2 = \frac{(1.776)^2}{2 \times 9.81} \]
\[ h_2 = \frac{3.16}{19.62} \]
\[ h_2 \approx 0.161 \, \text{m} \]

### Final Result
The height to which the ball will go up after it rebounds for the second time is approximately:
\[ h_2 \approx 0.161 \, \text{m} \text{ or } 16.1 \, \text{cm} \]