A body falling from a height of `10 m` rebounds from the hard floor . It

Options

A. `0.89`
B. `0.56`
C. `0.23`
D. `0.18`

Explanation

To solve the problem step by step, we need to find the coefficient of restitution between the ball and the floor after it rebounds from a height of 10 m, with the rebound height being reduced by 20%.

### Step 1: Calculate the rebound height
The initial height from which the body falls is 10 m. The rebound height is reduced by 20%.

\[ \text{Rebound height} = 10 \, \text{m} - (20\% \, \text{of} \, 10 \, \text{m}) = 10 \, \text{m} - 2 \, \text{m} = 8 \, \text{m} \]

### Step 2: Calculate the velocity just before impact
Using the equation of motion, we can find the velocity of the ball just before it hits the ground. The equation we will use is:

\[ v^2 = u^2 + 2gh \]

Where:
- \(u = 0\) (initial velocity)
- \(g = 9.81 \, \text{m/s}^2\) (acceleration due to gravity)
- \(h = 10 \, \text{m}\) (height)

Substituting the values:

\[ v^2 = 0 + 2 \cdot 9.81 \cdot 10 \]
\[ v^2 = 196.2 \]
\[ v = \sqrt{196.2} \approx 14.0 \, \text{m/s} \]

### Step 3: Calculate the velocity just after rebound
Using the same equation of motion for the rebound height of 8 m:

\[ u_b^2 = v^2 - 2gh' \]

Where:
- \(h' = 8 \, \text{m}\)

Substituting the values:

\[ u_b^2 = 0 + 2 \cdot 9.81 \cdot 8 \]
\[ u_b^2 = 156.96 \]
\[ u_b = \sqrt{156.96} \approx 12.5 \, \text{m/s} \]

### Step 4: Calculate the coefficient of restitution
The coefficient of restitution \(e\) is defined as:

\[ e = \frac{u_b}{v} \]

Substituting the values we found:

\[ e = \frac{12.5}{14.0} \]
\[ e \approx 0.893 \]

### Final Answer
The coefficient of restitution between the ball and the floor is approximately \(0.893\).

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