A ball is dropped onto the floor from a height of 10 m. It rebounds to a height of 5 m. If the ball was in contact with the floor for `0.01 s`, what was its average acceleration during contact ? (Take `g=10 ms^(-2)`)

Options

A. `2414 ms^(-2)`
B. `1735 ms^(-2)`
C. `3120 ms^(-2)`
D. `4105 ms^(-2)`

Explanation

To solve the problem of finding the average acceleration of the ball during its contact with the floor, we can follow these steps:

### Step 1: Determine the velocity just before impact
To find the velocity of the ball just before it hits the floor, we can use the equation of motion under gravity:

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

Where:
- \( v \) = final velocity (just before impact)
- \( u \) = initial velocity (0 m/s, since the ball is dropped)
- \( g \) = acceleration due to gravity (10 m/s²)
- \( h \) = height (10 m)

Substituting the values:

\[ v^2 = 0 + 2 \times 10 \times 10 \]
\[ v^2 = 200 \]
\[ v = \sqrt{200} = 14.14 \, \text{m/s} \]

### Step 2: Determine the velocity just after rebound
Next, we need to find the velocity of the ball just after it rebounds to a height of 5 m. We can use the same equation of motion:

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

Where:
- \( v' \) = final velocity (just after rebound)
- \( u' \) = initial velocity (0 m/s, at the peak of the rebound)
- \( g \) = acceleration due to gravity (10 m/s²)
- \( h' \) = height (5 m)

Substituting the values:

\[ v'^2 = 0 + 2 \times 10 \times 5 \]
\[ v'^2 = 100 \]
\[ v' = \sqrt{100} = 10 \, \text{m/s} \]

### Step 3: Calculate the change in velocity during contact
The change in velocity (\( \Delta v \)) during the contact with the floor can be calculated as:

\[ \Delta v = v' - (-v) = v' + v \]
\[ \Delta v = 10 \, \text{m/s} + 14.14 \, \text{m/s} = 24.14 \, \text{m/s} \]

### Step 4: Calculate the average acceleration
Average acceleration (\( a_{avg} \)) can be calculated using the formula:

\[ a_{avg} = \frac{\Delta v}{\Delta t} \]

Where \( \Delta t \) is the time of contact (0.01 s):

\[ a_{avg} = \frac{24.14 \, \text{m/s}}{0.01 \, \text{s}} = 2414 \, \text{m/s}^2 \]

### Final Answer
The average acceleration of the ball during contact with the floor is \( 2414 \, \text{m/s}^2 \).

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